SQL Queries (1)
1.Display the dept information from department table
select * from dept;
2.Display the details of all employees
select * from emp;
3.Display the name and job for all employees
select ename,job from emp;
4.Display name and salary for all employees
select ename,sal from emp;
5.Display employee number and total salary for each employee
select empno,sal+comm from emp;
6.Display employee name and annual salary for all employees
select empno,ename,12*sal+nvl(comm,0) annualsal from emp;
7.Display the names of all employees who are working in department number 10
select ename from emp where deptno = 10;
8.Display the names of all employees working as clerks and drawing a salary more than 3000
select ename from emp wher job = 'CLERK' and sal > 3000;
9.Display employee number and names for employees who earn commission
select empno,ename from emp where comm is not null and comm > 0;
10.Display names of employees who do not earn any commission
select empno,ename from emp where comm is null and comm= 0;
11.Display the names of employees who are working as clerk , salesman or analyst and drawing a salary more than 3000
select ename from emp where (job='CLERK' or job='SALESMAN' or job='ANALYST') and sal>3000;
12.Display the names of employees who are working in the company for the past 5 years
select ename from emp where sysdate - hiredate > 5*365;
13.Display the list of employees who have joined the company before 30 th june 90 or after 31 st dec 90
select * from emp where hiredate between '30-jun-1990' and '31-dec-1990';
14.Display current date
select sysdate from dual;
15.Display the list of users in your database (using log table)
select * from dba_users;
16.Display the names of all tables from the current user
select * from tab;
17.Display the name of the current user
show user;
18.Display the names of employees working in department number 10 or 20 or 40 or employees working as clerks , salesman or analyst
select ename from emp where deptno in (10,20,40) or job in ('CLERK','SALESMAN','ANALYST');
19.Display the names of employees whose name starts with alphabet s
select ename from emp where ename like 'S%';
20.Display employee name from employees whose name ends with alphabet S
select ename from emp where ename like '%S';
21.Display the names of employees whose names have sencond alphabet A in their names
select ename from emp where ename like '_S%';
22.Display the names of employees whose name is exactly five characters in length
select ename from emp where length(ename)=5;
(or )select ename from emp where ename like '_____';
23.Display the names of employees who are not working as managers
select * from emp minus (select * from emp where empno in (select mgr from emp));
(or ) select * from emp where empno not in (select mgr from emp where mgr is not null);
(or ) select * from emp e where empno not in (select mgr from emp where e.empno=mgr);
24.Display the names of employees who are not working as SALESMAN or CLERK or ANALYST
select job from emp where job not in ('CLERK','ANALYST','SALESMAN');
25.Display all rows from emp table. The system should wait after every screen full of information
set pause on;
26.Display the total number of employees working in the company
select count(*) from emp;
27.Display the total salary and total commission to all employees
select sum(sal), sum(nvl(comm,0)) from emp;
28.Display the maximum salary from emp table
select max(sal) from emp;
29.Display the minimum salary from emp table
select min(sal) from emp;
30.Display the average salary from emp table
select avg(sal) from emp;
31.Display the maximum salary being paid to CLERK
select max(sal) from emp where job='CLERK';
32.Display the maximum salary being paid in dept no 20
select max(sal) from emp where deptno=20;
33.Display the minimum salary being paid to any SALESMAN
select min(sal) from emp where job='SALESMAN';
34.Display the average salary drawn by managers.
select avg(sal) from emp where job='MANAGER';
35.Display the total salary drawn by analyst working in dept no 40
select sum(sal)+sum(nvl(comm,0)) from emp where deptno=40;
36.Display the names of employees in order of salary i.e. the name of the employee earning lowest salary shoud appear first
select ename from emp order by sal;
37.Display the names of employees in descending order of salary
select ename from emp order by sal desc;
38.Display the details from emp table in order of emp name
select ename from emp order by ename;
39.Display empnno,ename,deptno and sal. Sort the output first based on name and within name by deptno and witdhin deptno by sal;
select * from emp order by ename,deptno,sal;
40) Display the name of employees along with their annual salary(sal*12).
the name of the employee earning highest annual salary should appear first?
Ans:select ename,sal,sal*12 "Annual Salary" from emp order by "Annual Salary" desc;
41) Display name,salary,Hra,pf,da,TotalSalary for each employee.
The out put should be in the order of total salary ,hra 15% of salary ,DA 10% of salary .pf 5% salary Total Salary will be (salary+hra+da)-pf?
Ans: select ename,sal SA,sal*0.15 HRA,sal*0.10 DA,sal*5/100 PF, sal+(sal*0.15)+(sal*0.10)-(sal*.05) TOTALSALARY
from emp ORDER BY TOTALSALARY DESC;
42) Display Department numbers and total number of employees working in each Department?
Ans: select deptno,count(*) from tvsemp group by deptno;
43) Display the various jobs and total number of employees working in each job group?
Ans: select job,count(*) from tvsemp group by job;
44)Display department numbers and Total Salary for each Department?
Ans: select deptno,sum(sal) from tvsemp group by deptno;
45)Display department numbers and Maximum Salary from each Department?
Ans: select deptno,max(Sal) from tvsemp group by deptno;
46)Display various jobs and Total Salary for each job?
Ans: select job,sum(sal) from tvsemp group by job;
47)Display each job along with min of salary being paid in each job group?
Ans: select job ,min(sal) from tvsemp group by job;
48) Display the department Number with more than three employees in each department?
Ans: select deptno ,count(*) from tvsemp group by deptno having count(*)>3;
49) Display various jobs along with total salary for each of the job where total salary is greater than 40000?
Ans: select job,sum(sal) from tvsemp group by job having sum(SAl)>40000;
50) Display the various jobs along with total number of employees in each job.The
output should contain only those jobs with more than three employees?
Ans: select job,count(*) from tvsemp group by job having count(*)>3;
51) Display the name of employees who earn Highest Salary?
Ans: select ename, sal from tvsemp where sal>=(select max(sal) from tvsemp );
52) Display the employee Number and name for employee working as clerk and earning highest salary among the clerks?
Ans: select ename,empno from tvsemp where sal=(select max(sal) from tvsemp where job='CLERK') and job='CLERK' ;
53) Display the names of salesman who earns a salary more than the Highest Salary of the clerk?
Ans: select ename,sal from tvsemp where sal>(select max(sal) from tvsemp where job='CLERK') AND job='SALESMAN';
54) Display the names of clerks who earn a salary more than the lowest Salary of any salesman?
Ans: select ename,sal from tvsemp where sal>(select min(sal) from tvsemp where job='SALESMAN') and job='CLERK';
55) Display the names of employees who earn a salary more than that of jones or that of salary greater than that of scott?
Ans: select ename,sal from tvsemp where sal>all(select sal from tvsemp where ename='JONES' OR ename='SCOTT');
56) Display the names of employees who earn Highest salary in their respective departments?
Ans: select ename,sal,deptno from tvsemp where sal in (select max(sal) from tvsemp group by deptno);
57) Display the names of employees who earn Highest salaries in their respective job Groups?
Ans: select ename,job from tvsemp where sal in (select max(sal) from tvsemp group by job);
58) Display employee names who are working in Accounting department?
Ans: select e.ename,d.dname from emp e,dept d where e.deptno=d.deptno and d.dname='ACCOUNTING';
59) Display the employee names who are Working in Chicago?
Ans: select e.ename,d.loc from emp e,tvsdept d where e.deptno=d.deptno and d.loc='CHICAGO';
60) Display the job groups having Total Salary greater than the maximum salary for Managers?
Ans: select job ,sum(sal) from tvsemp group by job having sum(sal) >(select max(sal) from tvsemp where job='MANAGER');
61) Display the names of employees from department number 10 with salary greater than that of ANY employee working in other departments?
Ans: select ename,deptno from tvsemp where sal>any(select min(sal) from tvsemp where deptno!=10 group by deptno) and deptno=10 ;
62) Display the names of employees from department number 10 with salary greater than that of ALL employee working in other departments?
Ans: select ename,deptno from tvsemp where sal>all(select max(sal) from tvsemp where deptno!=10 group by deptno) and deptno=10 ;
63) Display the names of mployees in Upper Case?
Ans: select upper(ename) from tvsemp;
64) Display the names of employees in Lower Case?
Ans: select Lower(ename) from tvsemp;
65) Display the names of employees in Proper case?
Ans: select InitCap(ename)from tvsemp;
Q:66) Find the length of your name using Appropriate Function?
Ans: select lentgh('RAMA') from dual;
67) Display the length of all the employee names?
Ans: select length(ename) from tvsemp;
68) Display the name of employee Concatinate with Employee Number?
Ans: select ename||' '||empno from tvsemp;
69) Use appropriate function and extract 3 characters starting from 2 characters from the following string 'Oracle' i.e., the out put should be ac?
Ans: select substr('Oracle',3,2) from dual;
70) Find the first occurance of character a from the following string Computer Maintenance Corporation?
Ans: select lstr('Computer Maintenance Corporation','a' ) from dual;
71) Replace every occurance of alphabet A with B in the string .Alliens (Use Translate function)?
Ans: select translate('Alliens','A','B') from Dual;
72) Display the information from the employee table . where ever job Manager is found it should be displayed as Boss?
Ans: select ename ,replace(job,'MANAGER','BOSS') from tvsemp;
73) Display empno,ename,deptno from tvsemp table. Instead of display department numbers
display the related department name(Use decode function)?
Ans: select empno,ename,deptno,Decode(deptno,10,'ACCOUNTING'
,20,'RESEARCH',30,'SALES','OPERATIONS')DName from tvsemp;
74) Display your Age in Days?
Ans: select sysdate-to_date('30-jul-1977') from dual;
75) Display your Age in Months?
Ans: select months_between(sysdate,to_date('30-jul-1977')) from dual;
76) Display current date as 15th August Friday Nineteen Nienty Seven?
Ans: select To_char(sysdate,'ddth Month Day year') from dual;
77) Display the following output for each row from tvsemp table?
Ans: Q:78
78) Scott has joined the company on 13th August ninteen ninety?
Ans: select empno,ename,to_char(Hiredate,'Day ddth Month year') from tvsemp;
79) Find the nearest Saturday after Current date?
Ans: select next_day(sysdate,'Saturday') from dual;
80) Display the current time?
Ans: select To_Char(sysdate,'HH:MI:SS') from dual;
81) Display the date three months before the Current date?
Ans: select Add_months(sysdate,-3) from dual
82) Display the common jobs from department number 10 and 20?
Ans: select job from tvsemp where job in (select job from tvsemp where deptno=20) and deptno=10;
83) Display the jobs found in department 10 and 20 Eliminate duplicate jobs?
Ans: select Distinct job from tvsemp where deptno in(10,20);
84) Display the jobs which are unique to department 10?
Ans: select job from tvsemp where deptno=10;
85) Display the details of those employees who do not have any person working under him?
Ans: select empno,ename,job from tvsemp where empno not in (select mgr from tvsemp where mgr is not null );
86) Display the details of those employees who are in sales department and grade is 3?
Ans: select e.ename,d.dname,grade from emp e,dept d ,salgrade where e.deptno=d.deptno and dname='SALES' and grade=3;
87) Display thoes who are not managers?
Ans: select ename from tvsemp where job!='MANAGER';
88) Display those employees whose name contains not less than 4 characters?
Ans: select ename from tvsemp where length(ename)>=4
89) Display those department whose name start with"S" while location name ends with "K"?
Ans: select e.ename,d.loc from tvsemp e ,tvsdept d where d.loc like('%K') and ename like('S%')
90) Display those employees whose manager name is Jones?
Ans: select e.ename Superior,e1.ename Subordinate from tvsemp e,e1 where e.empno=e1.mgr and e.ename='JONES';
91) Display those employees whose salary is more than 3000 after giving 20% increment?
Ans: select ename,sal,(sal+(sal*0.20)) from tvsemp where (sal+(sal*0.20))>3000;
92) Display all employees with their department names?
Ans: select e.ename,d.dname from tvsemp e, tvsdept d where e.deptno=d.deptno
93) Display ename who are working in sales department?
Ans: select e.ename,d.dname from emp e,dept d where e.deptno=d.deptno and d.dname='SALES';
94) Display employee name,dept name,salary,and commission for those sal in between 2000
to 5000 while location is Chicago?
Ans: Select e.ename,d.dname,e.sal,e.comm from tvsemp e,dept d where e.deptno=d.deptno and sal between 2000 and 5000;
95) Display those employees whose salary is greater than his managers salary?
Ans: Select e.ename,e.sal,e1.ename,e1.sal from tvsemp e,e1 where e.mgr=e1.empno and e.sal>e1.sal;
96) Display those employees who are working in the same dept where his manager is work?
Ans: select e.ename,e.deptno,e1.ename,e1.deptno from tvsemp e,e1 where e.mgr=e1.empno and e.deptno=e1.deptno;
97) Display those employees who are not working under any Manager?
Ans: select ename from tvsemp where mgr is null;
98) Display the grade and employees name for the deptno 10 or 30 but grade is not 4 while
joined the company before 31-DEC-82?
Ans: select ename,grade,deptno,sal from tvsemp ,salgrade where ( grade,sal) in ( select grade,sal from salgrade,tvsemp where sal between losal and hisal) and grade!=4 and deptno in (10,30) and hiredate<'31-Dec-82';
99) Update the salary of each employee by 10% increment who are not eligible for commission?
Ans: update tvsemp set sal= (sal+(sal*0.10)) where comm is null;
100) Delete those employees who joined the company before 31-Dec-82 while their department Location is New York or Chicago?
Ans: select e.ename,e.hiredate,d.loc from tvsemp e,tvsdept d where
e.deptno=d.deptno and hiredate<'31-Dec-82' and d.loc in('NEW YORK','CHICAGO');
101) Display employee name ,job,deptname,loc for all who are working as manager?
Ans: select e.ename,e.job,d.dname,d.loc from tvsemp e,tvsdept d where e.deptno=d.deptno
and e.empno in (select mgr from tvsemp where mgr is not null);
102) Display those employees whose manager name is jones and also display their manager name?
Ans: select e.ename sub,e1.ename from tvsemp e,e1 where e.mgr=e1.empno and e1.ename='JONES';
103) Display name and salary of ford if his salary is equal to hisal of his grade?
Ans: select ename,grade,hisal,sal from emp,salgrade where ename='FORD' and sal=hisal;
(OR )select grade,sal,hisal from tvsemp,salgrade where ename='FORD' and sal between losal and hisal;
(OR )select ename,sal,hisal,grade from tvsemp,salgrade where ename='FORD'
and (grade,sal) in (select grade,hisal from salgrade,tvsemp where sal between losal and hisal);
104) Display employee name ,job,deptname,his manager name ,his grade and make an
under department wise?
Ans: select e.ename sub,e1.ename sup,e.job,d.dname ,grade from tvsemp e,e1,salgrade,tvsdept d where e.mgr=e1.empno and e.sal between losal and hisal and e.deptno=d.deptno group by d.deptno,e.ename,e1.ename,e.job,d.dname,grade;
(OR) select e.ename sub,e1.ename sup,e.job,d.dname ,grade from tvsemp e,e1,salgrade,tvsdept d where e.mgr=e1.empno and e.sal between losal and hisal and e.deptno=d.deptno
105) List out all the employee names ,job,salary,grade and deptname for every one in a company except 'CLERK' . Sort on salary display the highest salary?
Ans: select e.ename ,e.job,e.sal,d.dname ,grade from tvsemp e,salgrade,tvsdept d where (e.deptno=d.deptno and e.sal between losal and hisal ) order by e.sal desc
106) Display employee name,job abd his manager .Display also employees who are with out managers?
Ans: select e.ename ,e1.ename,e.job,e.sal,d.dname from tvsemp e,e1,tvsdept d where e.mgr=e1.empno(+) and e.deptno=d.deptno
107) Display Top 5 employee of a Company?
?
108) Display the names of those employees who are getting the highest salary?
Ans: select ename,sal from tvsemp where sal in (select max(sal) from tvsemp)
109) Display those employees whose salary is equal to average of maximum and minimum?
Ans: select * from tvsemp
where sal=(select (max(sal)+min(sal))/2 from tvsemp)
110) Select count of employees in each department where count >3?
Ans: select count(*) from tvsemp group by deptno having count(*)>3
111) Display dname where atleast three are working and display only deptname?
Ans: select d.dname from tvsdept d, tvsemp e where e.deptno=d.deptno group by d.dname having count(*)>3;
112) Display name of those managers name whose salary is more than average salary of Company?
Ans: select distinct e1.ename,e1.sal from tvsemp e,e1,dept d where e.deptno=d.deptno and e.mgr=e1.empno and e1.sal> (select avg(sal) from tvsemp);
113) Display those managers name whose salary is more than average salary salary of his employees?
Ans: select distinct e1.ename,e1.sal from tvsemp e,e1,dept d where e.deptno=d.deptno and e.mgr=e1.empno and e1.sal>any (select avg(sal) from tvsemp group by deptno);
114) Display employee name,sal,comm and netpay for those employees whose netpay is
greater than or equal to any other employee salary of the company?
Ans: select ename,sal,NVL(comm,0),sal+NVL(comm,0) from tvsemp where
sal+NVL(comm,0) >any (select e.sal from tvsemp e );
115) Display those employees whose salary is less than his manager but more than salary of other managers?
Ans: select e.ename sub,e.sal from tvsemp e,e1,tvsdept d where e.deptno=d.deptno and e.mgr=e1.empno
and e.salany (select e2.sal from tvsemp e2, e,tvsdept d1 where e.mgr=e2.empno and d1.deptno=e.deptno);
116) Display all employees names with total sal of company with each employee name?
Ans:
117) Find the last 5(least) employees of company?
Ans:
118) Find out the number of employees whose salary is greater than their managers salary?
Ans: select e.ename,e.sal,e1.ename,e1.sal from tvsemp e,e1,tvsdept d where e.deptno=d.deptno and e.mgr=e1.empno and e.sal>e1.sal
119) Display the manager who are not working under president but they are working under any other manager?
Ans: select e2.ename from emp e1,emp e2,emp e3 where e1.mgr=e2.empno and e2.mgr=e3.empno and e3.job!='PRESIDENT';
120) Delete those department where no employee working?
Ans: delete from tvsemp where empno is null;
121) Delete those records from emp table whose deptno not available in dept table?
Ans: delete from tvsemp e where e.deptno not in (select deptno from tvsdept)
122) Display those enames whose salary is out of grade available in salgrade table?
Ans: select empno,sal from tvsemp where sal<(select min(LOSAL) from salgrade )
OR sal>(select max(hisal) from salgrade)
123) Display employee name,sal,comm and whose netpay is greater than any othere in the company?
Ans: select ename,sal,comm,sal+comm from tvsemp where sal+comm>any (select sal+comm from tvsemp )
124) Display name of those employees who are going to retire 31-Dec-99 if maximum job period is 30 years?
Ans: select empno, hiredate,sysdate, to_char(sysdate,'yyyy') - to_char(hiredate,'yyyy')
from tvsemp where to_char(sysdate,'yyyy') - to_char(hiredate,'yyyy')=30
125) Display those employees whose salary is odd value?
Ans: select ename ,sal from tvsemp where mod(sal,2)!=0
126) Display those employees whose salary contains atleast 3 digits?
Ans: select ename,sal from tvsemp where length(sal)=3
127) Display those employees who joined in the company in the month of Dec?
Ans: Select empno,ename from tvsemp where trim(to_char(hiredate,'Mon'))=trim('DEC')
128) Display those employees whose name contains A?
Ans: select ename from tvsemp where ename like('%A%')
129) Display those employees whose deptno is available in salary?
Ans: select ename,sal from tvsemp where deptno in (select distinct sal from tvsemp);
130) Display those employees whose first 2 characters from hiredate - last 2 characters sal?
Ans: select empno,hiredate,sal from tvsemp where trim(substr(hiredate,1,2))=trim(substr(sal,-2,2));
(OR) select hiredate,sal from tvsemp where to_Char(hiredate,'dd')=trim(substr(sal,-2,2))
131) Display those employeess whose 10% of salary is equal to the year joining?
Ans: select ename ,sal,0.10*sal from tvsemp where 0.10*sal=trim(to_char(hiredate,'yy'))
132) Display those employees who are working in sales or research?
Ans: select e.ename from tvsemp e ,tvsdept d where e.deptno=d.deptno and d.dname in('SALES','RESEARCH');
133) Display the grade of jones?
Ans: select ename,grade from tvsemp,salgrade where ( grade,sal) =
(select grade,sal from salgrade,tvsemp where sal between losal and hisal and ename='JONES')
134) Display those employees who joined the company before 15th of the month?
Ans: select ename ,hiredate from tvsemp where hiredate<'15-Jul-02' and hiredate >='01-jul-02';
135) Display those employees who has joined before 15th of the month?
Ans: select ename ,hiredate from tvsemp where hiredate<'15-Jul-02'
136) Delete those records where no of employees in particular department is less than 3?
Ans: delete from tvsemp where deptno in (select deptno from tvsemp group by deptno having count(*) <3
137A) Delete those employeewho joined the company 10 years back from today?
Ans: delete from tvsemp where empno in (select empno from tvsemp
where to_char(sysdate,'yyyy')- to_char(hiredate,'yyyy')>=10)
137B) Display the deptname the number of characters of which is equal to no of employee
in any other department?
Ans:
138) Display the deptname where no employee is working?
Ans: select deptno from tvsemp where empno is null;
139) Display those employees who are working as manager?
Ans: select e2.ename from tvsemp e1,e2 where e1.mgr=e2.empno and e2.empno is not null
140) Count th number of employees who are working as managers (Using set opetrator)?
Ans: select d.dname from tvsdept d where length(d.dname) in (select count(*) from tvsemp e where e.deptno!=d.deptno group by e.deptno)
141) Display the name of the dept those employees who joined the company on the same date?
Ans: select a.ename,b.ename from tvsemp a,tvsemp b where a.hiredate=b.hiredate and a.empno!=b.empno
142) Display those employees whose grade is equal to any number of sal but not equal to first number of sal?
Ans: select ename,sal,grade ,substr(sal,grade,1) from tvsemp,salgrade where grade!=substr(sal,1,1) and grade = substr(sal,grade,1) and sal between losal and hisal
143) Count the no of employees working as manager using set operation?
Ans: Select count(empno) from tvsemp where empno in (select a.empno from tvsemp a intersect select b.mgr from tvsemp b)
144) Display the name of employees who joined the company on the same date?
Ans: select a.ename,b.ename from tvsemp a,tvsemp b where a.hiredate=b.hiredate and a.empno!=b.empno;
145) Display the manager who is having maximum number of employees working under him?
Ans: select e2.ename,count(*) from tvsemp e1,e2 where e1.mgr=e2.empno group by e2.ename Having count(*)=(select max(count(*)) from tvsemp e1,e2 where e1.mgr=e2.empno group by e2.ename)
146) List out the employee name and salary increased by 15% and express as whole number of Dollars?
Ans: select ename,sal,lpad(translate(sal,sal,((sal +(sal*0.15))/50)),5,'$') from tvsemp
147) Produce the output of the emptable "EMPLOYEE_AND JOB" for ename and job ?
Ans: select ename"EMPLOYEE_AND",job"JOB" FROM TVSEMP;
148) Lust of employees with hiredate in the format of 'June 4 1988'?
Ans: select ename,to_char(hiredate,'Month dd yyyy') from tvsemp;
149) print list of employees displaying 'Just salary' if more than 1500 if exactly 1500 display 'on taget' if less than 1500 display below 1500?
Ans: select ename,sal,
(
case when sal < 1500 then
'Below_Target'
when sal=1500 then
'On_Target'
when sal > 1500 then
'Above_Target'
else
'kkkkk'
end
)
from tvsemp
150) Which query to calc1ulate the length of time any employee has been with the company
Ans: select hiredate,to_char(hiredate,' HH:MI:SS') FROM tvsemp
151) Given a string of the format 'nn/nn' . Verify that the first and last 2 characters are numbers .And that the middle character is '/' Print the expressions 'Yes' IF valid 'NO' of not valid . Use the following values to test your solution'12/54',01/1a,'99/98'?
152) Employes hire on OR Before 15th of any month are paid on the last friday of that month
those hired after 15th are paid the last friday of th following month .print a list of employees .their hiredate and first pay date sort those who se salary contains first digit of their deptno?
Ans: select ename,hiredate, LAST_DAY ( next_day(hiredate,'Friday')),
(
case when to_char(hiredate,'dd') <=('15') then
LAST_DAY ( next_day(hiredate,'Friday'))
when to_char(hiredate,'dd')>('15') then
LAST_DAY( next_day(add_months(hiredate,1),'Friday'))
end
)
from tvsemp
153) Display those managers who are getting less than his employees salary?
Ans: select a.empno,a.ename ,a.sal,b.sal,b.empno,b.ename from tvsemp a, tvsemp b where a.mgr=b.empno and a.sal>b.sal
154) Print the details of employees who are subordinates to BLAKE?
Ans: select a.empno,a.ename ,b.ename from tvsemp a, tvsemp b where a.mgr=b.empno and b.ename='BLAKE'
151.Display those who working as manager using co related sub query
select * from emp where empno in (select mgr from emp);
152.Display those employees whose manager name is JONES and also with his manager name
select * from emp where mgr=(select empno from emp where ename='JONES') union select * from emp where empno = (select mgr from emp where ename='JONES');
153.Define variable representing the expressions used to calculate on employees total annual renumaration
define emp_ann_sal=(sal+nvl(comm,0))*.12;
154.Use the variable in a statement which finds all employees who can earn 30000 a year or more
select * from emp where &emp_ann_sal>30000;
155.Find out how many managers are there with out listing them
select count(*) from emp where empno in (select mgr from emp);
156.Find out the avg sal and avg total remuneration for each job type remember salesman earn commission
select job,avg(sal+nvl(comm,0)),sum(sal+nvl(comm,0)) from emp group by job;
157.Check whether all employees number are indeed unique
select count(empno) ,count(distinct(empno)) from emp having count(empno)=(count(distinct(empno));
158.List out the lowest paid employees working for each manager, exclude any groups where minsal is less than 1000 sort the output by sal
select e.ename,e.mgr,e.sal from emp e where sal in (select min(sal) from emp where mgr=e.mgr) and e.sal>1000 order by sal;
159.List ename,job,annual sal,depno,dname and grade who earn 30000 per year and who are not clerks
select e.ename,e.job,(e.sal+nvl(e.comm,0))*12,e.deptno,d.dname,s.grade from emp e,salgrade s,dept d where e.sal between s.losal and s.hisal and e.deptno=d.deptno and (e.sal+nvl(comm,0))*12 > 30000 and e.job<>'CLERK';
160.Find out th job that was falled in the first half of 1983 and the same job that was falled during the
same period on 1984
?
161.Find out the all employees who joined the company before their manager
select * from emp e where hiredate <(select hiredate from emp where empno=e.mgr);
162.List out the all employees by name and number along with their manager's name and number also display 'NO MANAGER' who has no manager
select e.empno,e.ename,m.empno Manager,m.ename ManagerName from emp e,emp m where e.mgr=m.empno
union
select empno,ename,mgr,'NO Manager' from emp where mgr is null;
163.Find out the employees who earned the highest sal in each job typed sort in descending sal order
select * from emp e where sal=(select max(sal) from emp where job=e.job);
164.Find out the employees who earned the min sal for their job in ascending order
select * from emp e where sal=(select min(sal) from emp where job=e.job) order by sal;
165.Find out the most recently hired employees in each dept order by hire date
select * from emp order by deptno,hiredate desc;
166.Display ename,sal and deptno for each employee who earn a sal greater than the avg of their department
order by deptno
select ename,sal,deptno from emp e where sal>(select avg(sal) from emp where deptno=e.deptno) order by deptno;
167.Display the department where there are no employees
select deptno,dname from dept where deptno not in (select distinct(deptno) from emp);
168.Display the dept no with highest annual remuneration bill as compensation
select deptno,sum(sal) from emp group by deptno having sum(sal)=(select max(sum(sal)) from emp group by deptno);
169.In which year did most people join the company. Display the year and number of employees
select count(*),to_char(hiredate,'yyyy') from emp group by to_char(hiredate,'yyyy');
170.Display avg sal figure for the dept
select deptno,avg(sal) from emp group by deptno;
171.Write a query of display against the row of the most recently hierd employee.display ename hire date
and column max date showing
select empno,hiredate from emp wher hiredate=(select max(hiredate) from emp);
172.Display employees who can earn more than lowest sal in dept no 30
select * from emp where sal > (select min(sal) from emp where deptno=30);
173.Find employees who can earn more than every employees in dept no 30
select * from emp where sal>(select max(sal) from emp where deptno=30);
select * from emp where sal>all(select sal from emp where deptno=30);
174.select dept name and deptno and sum of sal
break on deptno on dname;
select e.deptno,d.dname,sal from emp e,dept d where e.deptno=d.deptno order by e.deptno;
175.Find out avg sal and avg total remainders for each job type
176.Find all dept's which have more than 3 employees
select deptno from emp group by deptno having count(*)>3;
177.If the pay day is next Friday after 15th and 30th of every month. What is the next pay day from
their hire date for employee in emp table
178.If an employee is taken by you today in your organization and is a policy in your company to have a
review after 9 months the joined date (and of 1st of next month after 9 months) how many days from today
your employee has to wait for a review
?
179.Display employee name and his sal whose sal is greater than highest avg of deptno
?
180.Display the 10 th record of emp table (without using rowid)
?
181.Display the half of the enames in upper case and remaining lower case
select concat(upper(substr(ename,0,length(ename)/2),lower(substr(ename,length(ename)/2+1,length(ename)))) from
emp;
182.Display the 10th record of emp table without using group by and rowid
?
183.Delete the 10th record of emp table
?
184.Create a copy of emp table
create table emp1 as select * from emp;
185.select ename if ename exists more than once
select distinct(ename) from emp e where ename in (select ename from emp where e.empno<>empno);
186.Display all enames in reverse order
select ename from emp order by ename desc;
187.Display those employee whose joining of month and grade is equal
select empno,ename from emp e,salgrade s where e.sal between s.losal and s.hisal and to_char(hiredate,
'mm')=grade;
188.Display those employee whose joining date is available in deptno
select * from emp where to_char(hiredate,'dd') =deptno;
189.Display those employee name as follows A ALLEN, B BLAKE
select substr(ename,1,1)||''||ename from emp;
190.List out the employees ename,sal,pf from emp
select ename,sal,sal*15/100 pf from emp;
191.Display RSPS from emp without using updating,inserting
192.Create table emp with only one column empno
create table emp (empno number(5));
193.Add this column to emp table ename varchar2(20)
alter table emp add ename varchar2(20) not null;
194.OOPSI i forget to give the primary key constraint. Add it now
alter table emp add constraint emp_empno primary key (empno);
195.Now increase the length of ename column to 30 characters
alter table emp modify ename varchar2(30);
196.Add salary column to emp table
alter table emp add sal number(7,2);
197.I want to give a validation saying that sal can not be greater 10000(note give a name to this column)
alter table emp add constraint emp_sal_check check(sal<10000);
198.For the time being i have decided that i will not impose this validation. My boss has agreed to pay
more than 10000
alter table emp disable constraint emp_sal_check;
199.My boss has changed his mind. Now he doesn't want to pay more than 10000 So revoke that salary constraint
alter table emp enable constraint emp_sal_check;
200.Add column called as mgr to your emp table
alter table emp add mgr number(5);
201.Oh! This column should be related to empno, Give a command tdo add this constraint
Alter table emp add constraint emp_mgr foreign key (empno);
202.Add dept no column to your emp table
alter table emp add deptno number(3);
203.This deptno column should be related to deptno column of dept table
alter table emp1 add constraint emp1_deptno foreign key (deptno) references dept(deptno);
204.Create table called as new emp. Using single command create this table as well as to get data into
this table (use create table as)
create table newemp as select * from emp;
205.Create table called as newemp. This table should contain only empno,ename,dname
create table newemp as select empno,ename,dname from emp e,dept d where e.deptno=d.deptno;
206.Delete the rows of employees who are working in the company for more than 2 years
delete from emp where floor(sysdate-hiredate)>2*365;
207.Provides a commission to employees who are not earning any commission
select emp set comm=300 where comm is null;
208.If any employee has commission his commission should be incremented by 100% of his salary
update emp set comm=comm*10/100 where comm is not null;
209.Display employee name and department name for each employee
select ename,dname from emp e,dept d where e.deptno=d.deptno;
210.Display employee number,name and location of the department in which he is working
select empno,ename,loc from emp e,dept d where e.detpno=d.deptno;
211.Display ename,dname even if there no employees working in a particular department(use outer join)
select ename,dname from emp e,dept d where e.deptno(+)=d.deptno;
212.Display employee name and his manager name.
select e.ename,m.ename from emp e,emp m where e.mgr=m.empno;
213.Display the department name along with total salary in each department
select deptno,sum(sal) from emp group by deptno;
214.Display the department name and total number of employees in each department
select deptno,count(*) from emp group by deptno;
215.find out the top 3 sal
select top1 sal from(select distinct top3 sal from emp order by sal desc) FAQ
1. What is Log Switch? - The point at which ORACLE ends writing to one online redo log file and begins writing to another is called a log switch.
2. What is On-line Redo Log? - The On-line Redo Log is a set of tow or more on-line redo files that record all committed changes made to the database. Whenever a transaction is committed, the corresponding redo entries temporarily stores in redo log buffers of the SGA are written to an on-line redo log file by the background process LGWR. The on-line redo log files are used in cyclical fashion.
3. Which parameter specified in the DEFAULT STORAGE clause of CREATE TABLESPACE cannot be altered after creating the tablespace? - All the default storage parameters defined for the tablespace can be changed using the ALTER TABLESPACE command. When objects are created their INITIAL and MINEXTENS values cannot be changed.
4. What are the steps involved in Database Startup? - Start an instance, Mount the Database and Open the Database.
5. What are the steps involved in Instance Recovery? - Rolling forward to recover data that has not been recorded in data files, yet has been recorded in the on-line redo log, including the contents of rollback segments. Rolling back transactions that have been explicitly rolled back or have not been committed as indicated by the rollback segments regenerated in step a. Releasing any resources (locks) held by transactions in process at the time of the failure. Resolving any pending distributed transactions undergoing a two-phase commit at the time of the instance failure.
6. Can Full Backup be performed when the database is open? - No.
7. What are the different modes of mounting a Database with the Parallel Server? - Exclusive Mode If the first instance that mounts a database does so in exclusive mode, only that Instance can mount the database. Parallel Mode If the first instance that mounts a database is started in parallel mode, other instances that are started in parallel mode can also mount the database.
8. What are the advantages of operating a database in ARCHIVELOG mode over operating it in NO ARCHIVELOG mode? - Complete database recovery from disk failure is possible only in ARCHIVELOG mode. Online database backup is possible only in ARCHIVELOG mode.
9. What are the steps involved in Database Shutdown? - Close the Database, Dismount the Database and Shutdown the Instance.
10. What is Archived Redo Log? - Archived Redo Log consists of Redo Log files that have archived before being reused.
11. What is Restricted Mode of Instance Startup? - An instance can be started in (or later altered to be in) restricted mode so that when the database is open connections are limited only to those whose user accounts have been granted the RESTRICTED SESSION system privilege.
12. What is Partial Backup? - A Partial Backup is any operating system backup short of a full backup, taken while the database is open or shut down.
13. What is Mirrored on-line Redo Log? - A mirrored on-line redo log consists of copies of on-line redo log files physically located on separate disks, changes made to one member of the group are made to all members.
14. What is Full Backup? - A full backup is an operating system backup of all data files, on- line redo log files and control file that constitute ORACLE database and the parameter.
15. Can a View based on another View? - Yes.
16. Can a Tablespace hold objects from different Schemes? - Yes.
17. Can objects of the same Schema reside in different tablespaces? - Yes.
18. What is the use of Control File? - When an instance of an ORACLE database is started, its control file is used to identify the database and redo log files that must be opened for database operation to proceed. It is also used in database recovery.
19. Do View contain Data? - Views do not contain or store data.
20. What are the Referential actions supported by FOREIGN KEY integrity constraint? - UPDATE and DELETE Restrict - A referential integrity rule that disallows the update or deletion of referenced data. DELETE Cascade - When a referenced row is deleted all associated dependent rows are deleted.
21. What are the type of Synonyms? - There are two types of Synonyms Private and Public
22. What is a Redo Log? - The set of Redo Log files YSDATE,UID,USER or USERENV SQL functions, or the pseudo columns LEVEL or ROWNUM.
23. What is an Index Segment? - Each Index has an Index segment that stores all of its data.
24. Explain the relationship among Database, Tablespace and Data file.? - Each databases logically divided into one or more tablespaces one or more data files are explicitly created for each tablespace
25. What are the different type of Segments? - Data Segment, Index Segment, Rollback Segment and Temporary Segment.
26. What are Clusters? - Clusters are groups of one or more tables physically stores together to share common columns and are often used together.
27. What is an Integrity Constrains? - An integrity constraint is a declarative way to define a business rule for a column of a table.
28. What is an Index? - An Index is an optional structure associated with a table to have direct access to rows, which can be created to increase the performance of data retrieval. Index can be created on one or more columns of a table.
29. What is an Extent? - An Extent is a specific number of contiguous data blocks, obtained in a single allocation, and used to store a specific type of information.
30. What is a View? - A view is a virtual table. Every view has a Query attached to it. (The Query is a SELECT statement that identifies the columns and rows of the table(s) the view uses.)
31. What is Table? - A table is the basic unit of data storage in an ORACLE database. The tables of a database hold all of the user accessible data. Table data is stored in rows and columns.
32. What is a Synonym? - A synonym is an alias for a table, view, sequence or program unit.
33. What is a Sequence? - A sequence generates a serial list of unique numbers for numerical columns of a database’s tables.
34. What is a Segment? - A segment is a set of extents allocated for a certain logical structure.
35. What is schema? - A schema is collection of database objects of a User.
36. Describe Referential Integrity? - A rule defined on a column (or set of columns) in one table that allows the insert or update of a row only if the value for the column or set of columns (the dependent value) matches a value in a column of a related table (the referenced value). It also specifies the type of data manipulation allowed on referenced data and the action to be performed on dependent data as a result of any action on referenced data.
37. What is Hash Cluster? - A row is stored in a hash cluster based on the result of applying a hash function to the row’s cluster key value. All rows with the same hash key value are stores together on disk.
38. What is a Private Synonyms? - A Private Synonyms can be accessed only by the owner.
39. What is Database Link? - A database link is a named object that describes a “path” from one database to another.
40. What is a Tablespace? - A database is divided into Logical Storage Unit called tablespaces. A tablespace is used to grouped related logical structures together
41. What is Rollback Segment? - A Database contains one or more Rollback Segments to temporarily store “undo” information.
42. What are the Characteristics of Data Files? - A data file can be associated with only one database. Once created a data file can’t change size. One or more data files form a logical unit of database storage called a tablespace.
43. How to define Data Block size? - A data block size is specified for each ORACLE database when the database is created. A database users and allocated free database space in ORACLE datablocks. Block size is specified in INIT.ORA file and can’t be changed latter.
44. What does a Control file Contain? - A Control file records the physical structure of the database. It contains the following information. Database Name Names and locations of a database’s files and redolog files. Time stamp of database creation.
45. What is difference between UNIQUE constraint and PRIMARY KEY constraint? - A column defined as UNIQUE can contain Nulls while a column defined as PRIMARY KEY can’t contain Nulls. 47.What is Index Cluster? - A Cluster with an index on the Cluster Key 48.When does a Transaction end? - When it is committed or Rollbacked.
46. What is the effect of setting the value “ALL_ROWS” for OPTIMIZER_GOAL parameter of the ALTER SESSION command? - What are the factors that affect OPTIMIZER in choosing an Optimization approach? - Answer The OPTIMIZER_MODE initialization parameter Statistics in the Data Dictionary the OPTIMIZER_GOAL parameter of the ALTER SESSION command hints in the statement.
47. What is the effect of setting the value “CHOOSE” for OPTIMIZER_GOAL, parameter of the ALTER SESSION Command? - The Optimizer chooses Cost_based approach and optimizes with the goal of best throughput if statistics for atleast one of the tables accessed by the SQL statement exist in the data dictionary. Otherwise the OPTIMIZER chooses RULE_based approach.
48. What is the function of Optimizer? - The goal of the optimizer is to choose the most efficient way to execute a SQL statement.
49. What is Execution Plan? - The combinations of the steps the optimizer chooses to execute a statement is called an execution plan.
50. What are the different approaches used by Optimizer in choosing an execution plan? - Rule-based and Cost-based.
1. How can I retrieve data from a database using SQL?
The SELECT command is the most commonly used command in SQL. It allows database users to retrieve the specific information they desire from an operational database.
2. How do I create a new database or a new database table?
SQL provides the CREATE DATABASE and CREATE TABLE commands to add new databases and tables, respectively to your database. These commands provide a highly flexible syntax allowing you to create tables and databases that meet your specific business requirements.
3. How do I add data to a database?
The INSERT command in SQL is used to add records to an existing table.
4. How do I delete some or all of a database table?
Oftentimes, it becomes necessary to remove obsolete information from a relational database. Fortunately, Structured Query Language provides a flexible DELETE command that can be used to remove some or all of the information stored within a table.
5. What is a NULL value?
NULL is the value used to represent an unknown piece of data. Databases treat NULL values in a special way, depending upon the type of operation that it is used in. When a NULL value appears as an operand to an AND operation, the operation’s value is FALSE if the other operand is FALSE (there is no way the expression could be TRUE with one FALSE operand). On the other hand, the result is NULL (unknown) if the other operand is either TRUE or NULL (
6. How can I combine data from multiple database tables?
SQL join statements allow you to combine data from two or more tables in your query results. Learn how to leverage this powerful technology to supercharge your database queries.
7. Can I join a table to itself?
Yes! You can use a self-join to simplify nested SQL queries where the inner and outer queries reference the same table. These joins allow you to retrieve related records from the same table.
8. How can I summarize data contained within a database table?
SQL provides aggregate functions to assist with the summarization of large volumes of data. The SUM function is used within a SELECT statement and returns the total of a series of values. The AVG function works in a similar manner to provide the mathematical average of a series of values. SQL provides the COUNT function to retrieve the number of records in a table that meet given criteria. The MAX() function returns the largest value in a given data series while the MIN() function returns the smallest value.
9. How can I group summarized data?
You may use basic SQL queries to retrieve data from a database but this often doesn’t provide enough intelligence to meet business requirements. SQL also provides you with the ability to group query results based upon row-level attributes in order to apply aggregate functions using the GROUP BY clause.
10. How can I restrict access to data contained within an SQL database?
SQL databases provide administrators with a role-based access control system. In this schema, administrators create user accounts for each individual database user and then assign that user to one or more database roles that describe a way the user is allowed to interact with the database. Finally, the administrator grants specific permissions to the role to allow role members to carry out the desired actions. Users are implicitly denied any access that they are not explicitly granted.
SAP BUSINESSOBJECTS Designer Overview
This Section describes the basic operations you perform in Designer to create, modify, and update universes. The following topics are covered:
1) Starting Designer, 2)Importing a universe, 3)Opening a universe,4) Exporting a universe, 5)Saving a universe, 6)Creating a universe,7) Setting universe parameters,8) Using the Designer user interface 9)Using Find and Replace, 10)Organizing the table display,11) Selecting schema display options12)Printing a universe
Note: $INSTALLDIR variable in this guide
In this guide the variable $INSTALLDIR is the install root path for the data access files used by Designer and Web Intelligence. This is the Business Objects installation path with the operating system sub directory that contains the Designer executable and the data access drivers.
Under Windows$INSTALLDIR = \\…\Business Objects\BusinessObjects
Enterprise 11\win32_x86.
For example C:\Program Files\Business Objects\SAP BusinessObjects Enterprise
11\win32_x86.
Starting Designer Designer can only be used with a Central Management System (CMS)
repository. You must log in to the repository before starting Designer.
If you are starting Designer for the first time and want to work on an existing universe, you need to open the universe directly first, save it with a secure connection and export it to the repository. You then import the universe to make updates and export updated versions. This ensures that the CMS and the local universe versions are synchronized.
Creating a new universe
The following procedure describes how you can create a new universe from scratch by defining universe parameters then saving the universe. The procedure provides an overview of all the pages available from the Parameters dialog box.
For more detailed information on each step you should refer to the respective section for the parameter in this chapter.
Defining all the parameters at universe creation may not be necessary. You must select a connection, but you can accept the default values for other parameters, and then modify them as appropriate when necessary.
Creating a new universe from scratch
To create a new universe from scratch:
1. Select File > New.
The Universe parameters dialog box opens to the Definition page.
• Type a name and description for the universe.
• Select a connection from the Connection drop-down list box.(Or)
• Click the New button if you want to define a new connection that isnot listed in the drop-down list.
2. Click the Summary tab.
The Summary page appears.
• Type universe information in the Comments box.
3. Click the Strategies tab.
The Strategies page appears. It displays the strategies available for your connected data source.
• Select a strategy from each of the Objects, Joins, and Tables dropdown list boxes.
Depending on the RDBMS for the connection, there can be more than one strategy available from each drop-down list box.
4. Click the Controls tab.
The Controls page appears.
• Select or clear check boxes in the Query Limits group box.
• Enter values for the check boxes that you select.
5. Click the SQL tab.
The SQL page appears. See the “Indicating SQL restrictions” on page 62 for information on this page.
• Select or clear check boxes as appropriate.
6. Click the Links tab, if you want to link the new universe with an existing universe.
The Links page appears.
• Click the Add Link button to select a universe to link with the new universe.
7. Click the Parameters tab.
The Parameters page appears. It lists SQLparameters that can be set to optimize SQL generation.
8. Click OK.
The universe and structure panes open up in Designer
9. Select File > Save.
• Type a name for the universe file.
This Section describes the basic operations you perform in Designer to create, modify, and update universes. The following topics are covered:
1) Starting Designer, 2)Importing a universe, 3)Opening a universe,4) Exporting a universe, 5)Saving a universe, 6)Creating a universe,7) Setting universe parameters,8) Using the Designer user interface 9)Using Find and Replace, 10)Organizing the table display,11) Selecting schema display options12)Printing a universe
Note: $INSTALLDIR variable in this guide
In this guide the variable $INSTALLDIR is the install root path for the data access files used by Designer and Web Intelligence. This is the Business Objects installation path with the operating system sub directory that contains the Designer executable and the data access drivers.
Under Windows$INSTALLDIR = \\…\Business Objects\BusinessObjects
Enterprise 11\win32_x86.
For example C:\Program Files\Business Objects\SAP BusinessObjects Enterprise
11\win32_x86.
Starting Designer Designer can only be used with a Central Management System (CMS)
repository. You must log in to the repository before starting Designer.
If you are starting Designer for the first time and want to work on an existing universe, you need to open the universe directly first, save it with a secure connection and export it to the repository. You then import the universe to make updates and export updated versions. This ensures that the CMS and the local universe versions are synchronized.
Creating a new universe
The following procedure describes how you can create a new universe from scratch by defining universe parameters then saving the universe. The procedure provides an overview of all the pages available from the Parameters dialog box.
For more detailed information on each step you should refer to the respective section for the parameter in this chapter.
Defining all the parameters at universe creation may not be necessary. You must select a connection, but you can accept the default values for other parameters, and then modify them as appropriate when necessary.
Creating a new universe from scratch
To create a new universe from scratch:
1. Select File > New.
The Universe parameters dialog box opens to the Definition page.
• Type a name and description for the universe.
• Select a connection from the Connection drop-down list box.(Or)
• Click the New button if you want to define a new connection that isnot listed in the drop-down list.
2. Click the Summary tab.
The Summary page appears.
• Type universe information in the Comments box.
3. Click the Strategies tab.
The Strategies page appears. It displays the strategies available for your connected data source.
• Select a strategy from each of the Objects, Joins, and Tables dropdown list boxes.
Depending on the RDBMS for the connection, there can be more than one strategy available from each drop-down list box.
4. Click the Controls tab.
The Controls page appears.
• Select or clear check boxes in the Query Limits group box.
• Enter values for the check boxes that you select.
5. Click the SQL tab.
The SQL page appears. See the “Indicating SQL restrictions” on page 62 for information on this page.
• Select or clear check boxes as appropriate.
6. Click the Links tab, if you want to link the new universe with an existing universe.
The Links page appears.
• Click the Add Link button to select a universe to link with the new universe.
7. Click the Parameters tab.
The Parameters page appears. It lists SQLparameters that can be set to optimize SQL generation.
8. Click OK.
The universe and structure panes open up in Designer
9. Select File > Save.
• Type a name for the universe file.
Setting universe parameters
You can set universe parameters for the following purposes:
• Identifying the universe
• Defining and editing connections
• Viewing and entering summary information
• Selecting strategies
• Indicating resource controls
• Indicating SQL restrictions
• Indicating options for linked universes
• Setting SQL generation parameters
Each type of parameter is contained on a page in the Parameters dialog box
(File > Parameters).Each group of parameters is described in its respective section below.
Modifying universe identification parameters
To modify universe identification parameters:
1. Select File > Parameters. (Or)
Click the Universe Parameters button in the toolbar.
The Universe Parameters dialog box opens to the Definition page.
2. Type a name and a description.
3. Select a connection from the Connection drop-down list box.
4. Click the Test button to verify that the connection is valid.
If you receive a message informing you that the server is not responding, the connection is not valid. You can correct connection parameters by clicking the Edit button and editing connection properties. If the error persists, refer to the section of the RDBMS documentation relating to
error messages.
5. Click OK.
Defining and editing connections
A connection is a named set of parameters that defines how a Business Objects application accesses data in a database file. A connection links Web Intelligence to your middleware. You must have a connection to access data. modify, delete, or replace the connection at any time.
Note: See the Data Access Guide for complete information on creating, modifying,
and optimizing connections
You can create a new connection from the Definition page of the Universe Parameters dialog box (File > Parameters > Definition). You create a new connection when there is not an existing connection appropriate to the current universe. You can also edit the properties for a connection from the Definition page.
You can view all connections available to a universe from the Connections list
(Tools > Connections). You can delete, edit, and create new connections from this page.
A connection contains three elements:
• Data Access driver
• Connection and login parameters
• Connection type
Each element is described in the following sections:
Data Access driver
A Data Access driver is the software layer that connects a universe to your middleware.
Data Access drivers are shipped with SAP SAP BusinessObjects products. There is a
Data Access driver for each supported middleware. When you install
Designer, your Data Access key determines which Data Access drivers are installed.
When you create a new connection, you select the appropriate Data Access driver for the RDBMS middleware that you use to connect to the target RDBMS
You can set universe parameters for the following purposes:
• Identifying the universe
• Defining and editing connections
• Viewing and entering summary information
• Selecting strategies
• Indicating resource controls
• Indicating SQL restrictions
• Indicating options for linked universes
• Setting SQL generation parameters
Each type of parameter is contained on a page in the Parameters dialog box
(File > Parameters).Each group of parameters is described in its respective section below.
Modifying universe identification parameters
To modify universe identification parameters:
1. Select File > Parameters. (Or)
Click the Universe Parameters button in the toolbar.
The Universe Parameters dialog box opens to the Definition page.
2. Type a name and a description.
3. Select a connection from the Connection drop-down list box.
4. Click the Test button to verify that the connection is valid.
If you receive a message informing you that the server is not responding, the connection is not valid. You can correct connection parameters by clicking the Edit button and editing connection properties. If the error persists, refer to the section of the RDBMS documentation relating to
error messages.
5. Click OK.
Defining and editing connections
A connection is a named set of parameters that defines how a Business Objects application accesses data in a database file. A connection links Web Intelligence to your middleware. You must have a connection to access data. modify, delete, or replace the connection at any time.
Note: See the Data Access Guide for complete information on creating, modifying,
and optimizing connections
You can create a new connection from the Definition page of the Universe Parameters dialog box (File > Parameters > Definition). You create a new connection when there is not an existing connection appropriate to the current universe. You can also edit the properties for a connection from the Definition page.
You can view all connections available to a universe from the Connections list
(Tools > Connections). You can delete, edit, and create new connections from this page.
A connection contains three elements:
• Data Access driver
• Connection and login parameters
• Connection type
Each element is described in the following sections:
Data Access driver
A Data Access driver is the software layer that connects a universe to your middleware.
Data Access drivers are shipped with SAP SAP BusinessObjects products. There is a
Data Access driver for each supported middleware. When you install
Designer, your Data Access key determines which Data Access drivers are installed.
When you create a new connection, you select the appropriate Data Access driver for the RDBMS middleware that you use to connect to the target RDBMS
Connection type
The type of connection determines who can use the connection to access data. Designer automatically stores all the connections that you create during a work session. The next time you launch a session, these connections will be available to you.
You can create three types of connections with Designer:
• Personal
• Shared
• Secured
Each connection type is described as follows:
Personal connections
Restricts access to data to the universe creator and the computer on which it was created. Connection parameters are stored in the PDAC.LSI file located in the LSI folder in the SAP SAP Business Objects installation path. An example of this path is shown below.
C:\Program Files\Business Objects\SAP Business Objects Enterprise
11\win32_x86\pdac.lsi
These parameters are static and cannot be updated.
Personal connections are unsecured in terms of SAP SAP Business Objects products security.
You do not use personal connections to distribute universes. You could use personal connections to access personal data on a local machine.
The type of connection determines who can use the connection to access data. Designer automatically stores all the connections that you create during a work session. The next time you launch a session, these connections will be available to you.
You can create three types of connections with Designer:
• Personal
• Shared
• Secured
Each connection type is described as follows:
Personal connections
Restricts access to data to the universe creator and the computer on which it was created. Connection parameters are stored in the PDAC.LSI file located in the LSI folder in the SAP SAP Business Objects installation path. An example of this path is shown below.
C:\Program Files\Business Objects\SAP Business Objects Enterprise
11\win32_x86\pdac.lsi
These parameters are static and cannot be updated.
Personal connections are unsecured in terms of SAP SAP Business Objects products security.
You do not use personal connections to distribute universes. You could use personal connections to access personal data on a local machine.
Shared connectionsAllows access to data for all users. These connections are unsecured in terms of SAP SAP Business Objects products security.
Connection parameters are stored in the SDAC.LSI file located in the LSI folder in the SAP SAP Business Objects installation path. An example of this path is shown below.
C:\Program Files\Business Objects\SAP Business Objects Enterprise
11\win32_x86\sdac.lsi
If the SDAC.SSI file is stored locally, only users having access to the local machine (through a mapped drive), can use the shared connections .Shared connections can be useful in a universe testing environment.
Secured connections
• Centralizes and controls access to data. It is the safest type of connection, and should used be to protect access to sensitive data.
• You can create secured connections with Designer.
• You must use secured connections if you want to distribute universes through the CMS.
• Secured connections can be used and updated at any time
Connection parameters are stored in the SDAC.LSI file located in the LSI folder in the SAP SAP Business Objects installation path. An example of this path is shown below.
C:\Program Files\Business Objects\SAP Business Objects Enterprise
11\win32_x86\sdac.lsi
If the SDAC.SSI file is stored locally, only users having access to the local machine (through a mapped drive), can use the shared connections .Shared connections can be useful in a universe testing environment.
Secured connections
• Centralizes and controls access to data. It is the safest type of connection, and should used be to protect access to sensitive data.
• You can create secured connections with Designer.
• You must use secured connections if you want to distribute universes through the CMS.
• Secured connections can be used and updated at any time
Usually there are three kinds of backups
They are 1. Complete or Full backup 2. Differential Backup 3. Transaction Log Backup
Complete Backup: A complete Backup has to be taken regardless or the importance of database, usually at least once a month depends on the environment you are working. I guess you know how to take a basic full backup. To take a full backup the database shouldn't be necessarily in FULL recovery mode, it can be in simple mode as well. In our environment we take a full backup usually once a week.
Transaction Log Backup: Before going into this details, try to understand what is the significance of this backup, it basically has the transactions like insertions, deletions or any DDL operations you have done on the database, so this is a very important part of your DB growth. Say suppose if there are more transactions going on your DB(insertions, deletions, updates) then you need to backup transactions logs frequently, if not your DISK will get out of space with those transactions, also many DBA keeps a monitoring on their DISK space. In our Case some DB"S which are very critical we usually have them log shipped over to our DISASTER RECOVERY SITE( DR SITE) which is in DENVER, and for most of the other databases, we take TL backup once a day because we can afford to lose data within a day (will understand more about it later section).
Differential Backup : going by the name, it tis the difference of the TL backups, this is used in mission critical applications, where in FULL is taken once aweek usually, differential 2 or 3 times a week and transactions log every 4 or 6 hrs (aproximately just to give you an idea)
Failure Scenario: Say you had a system failure and you need to recover your DB, the 1st step you want to see is if you have a Latest FULL backup of it. Going by our scenario, consider the assumptions below
Full Backup: praveen2409.bak (backup file on 24th of this month)
Transaction: TL2409.trn, TL2509.trn, TL2609.trn, TL2709.trn (transaction log backup daily from 22nd till date)
Now on 28th your system crashes at 3 pm afternoon on 29th, so now you need to recover the DB from the complete backup and the 4 transactions log backups you have,
First, apply the recovery of FULL backup with no recovery option, and then in sequential order do the recover for the rest of transaction logs (i.e TL2409.trn, TL2509.trn.....) For the final file , you need to mention the With Recovery option which would recover the DB completely till 27th. Now in this case you may lose 28th data till 3 pm, if this process is acceptable by your environment.
Scenario 2:
Now you say, no i cant afford to lose this much of data, then you may want to increase you TL frequency (maybe once in 4 hours)
Now again the recovery of a DB in this process is a "pain in the ass" as applying a full backup followed by 24 TL backup's (6 per day * 4 days) considering system fails after 4 days of full backup. In this case you need the differential backup where this would take the difference of the TL backup and in this case you may need to apply the latest differential Backup.
So the full backup is once a week, differential is once a day and TL is once in 4 hours, so while recovering apply FULL followed by differential till 27th and transaction log after that till the last one before 3pm where system crashed. Say if the latest TL backup was at 1 pm, so in this case you’ll lose at the most 2 hrs of data.
STORED PROCEDURE
Stored procedures are essentially functions that you can create in the database and reuse.
SYNTAX CREATE PROCEDURE procedure_name
BENEFITS OF STORED PROCEDURE
· Speed - Stored procedures are pre-compiled, so the execution plan doesn't have to be figured out each time they're called.
· Code reuse and abstraction - Stored procedures often involve complex code. It's nice that this code doesn't have to be rewritten over and over again and that even entry-level SQL programmers can make use of it with a simple stored procedure call.
· Security - Permissions can be granted for stored procedures while being restricted for the underlying tables. This allows the DBA to provide a method for SQL programmers and report writers to access and/or manipulate data in a safe way.
· Reduced traffic between client and server - Since the bulk of the query is already stored on the server and only the relatively short stored procedure call has to get sent, traffic to the server is decreased.
TRIGGER
A TRIGGER is a special type of stored procedure, which is 'fired' automatically when the data in a specified table is modified. It is invoked when an INSERT, UPDATE, or DELETE action is performed on a table.
What is RDBMS?
Relational Data Base Management Systems (RDBMS) are database management systems that maintain data records and indices in tables. Relationships may be created and maintained across and among the data and tables. In a relational database, relationships between data items are expressed by means of tables. Interdependencies among these tables are expressed by data values rather than by pointers. This allows a high degree of data independence. An RDBMS has the capability to recombine the data items from different files, providing powerful tools for data usage.
Q) Those who are coming under excellent need to add 0.05 % bonus
SELECT SAL+(SAL*0.05) FROM EMP WHERE SAL>1500;
Or
SELECT ROUND(SAL+(SAL*0.05)) FROM EMP WHERE SAL>1500;
alter table ramanaidu add(emp_doj date);
insert into ramanaidu values('&slno','&ename','&job','&empid','&emp_add','&emp_mob','&emp_email')
create table ramanaidu(slno number(5),Ename varchar2(10),job varchar2(10),empid number(4),emp_add varchar2(30),emp_mob number(10),emp_email varchar2(10));
update ramanaidu set emp_doj='05-APR-2011' where slno=1;
update kolli set emp_mob=9972851171 where emp_mob=8050834789;
update ramanaidu set emp_doj='01-Jan-2011' where job='QCA'
alter table ramanaidu modify emp_email varchar2(20);
DDL – Data Defination Language – Create, Alter,rename,drop
DML- Data Manipulation Language – Insert,Update,delete
DQL – Data Query Lanugage or Data Retrival Language – only select command
DCL – Data Control Language – grant, revoke – User Permission and User Permission cancle
Drop table table name;
Rename table name to new table name – rename Ramanaidu to kolli;
Alter table kolli drop column emp_email; --------- for deleting column permanently
Alter table kolli rename column ename to empname; - for column name eding purpose
Create table ramanaidu as select * from kolli; - Copying table as it is
Create table EMP as select * from emp1;
ALTER TABLE KOLLI RENAME COLUMN NAME TO ENAME – CHANGING COLUMN NAME;
Delete table Name - it will be deleted data (records) structed will be there
Drop table table name - it will be delete entire table and structure
Create table table name (table column name data type)
Select sysdate from dual; - displaying system date – dummy column
Select 5+3 from dual;