Tuesday, September 29, 2009

Resetting High Water Mark in Oracle10g

I would like to write how to reset HWM in oracle10g. Prior to Oracle10g, resetting high water mark is painful procedure in busy environment. Oracle made our life easy to reset the High Water Mark in oracle10g. I am not going to discuss about what is HWM. Since i already discussed this in another topic. Please click to read about what is HWM and what are the various options available to reset the HMW.

We have traditional ways to reset the HWM prior to Oracle10g. We might need Table downtime when we use below traditional methods....

1. Imp/exp
2. Alter tablespace move
3. truncate and insert
4. user dbms_redefinition package to copy the table

Okay... Let us talk about resetting HWM in Oracle10g. The tablespace should be ASSM(Automatic segment space Management) enabled to leverge this feature. In oracle10g, we do not need table downtime to reset the HWM. It would be easy to reset the HWM in 24/7 environment.

What does Oracle do while using Oracle10g feature to reset the HMW?

Oracle split this process as two phase. Let me explain what is happening in each phase.

Phase I. Oracle move the rows which are located in the middle or at the end of a segment further more down to the beginning of the segment and make the segment more compact. This shrinking process is kind of delete and insert. But it is not really!!!. This process is moving the data row by row. It acquires a row level lock when the row is moved down to the begining of the segment. The corresponding index data will be handled like any other row level DML. So we do not need to worry about rebuilding the indexes for the row. Also row level lock will happen for very short moment. Before we start this phase, we need to enable row movement. Here is the command to complete the phase I. Here i am using the table called bookings.

SQL> alter table bookings enable row movement;

Table altered.

SQL> alter table bookings shrink space compact;

Table altered

Phase II This step will reset the high water mark. This will acquire table level lock for very short moment while resetting the HWM. Here is the command to accomplish this task.

SQL> alter table bookings shrink space;

Table altered

SQL> alter table bookings shrink space cascade; (it is for all dependent objects as well)

Table altered

If we want to reset the HWM in one go, then below command will accomplish the task. The below command moves the rows and reset the HWM.

SQL> alter table bookings shrink space;

Table altered

What are the advantages of using Oracle10g new feature to reset the HWM?

There are serveral advantages over traditional methods. Let me list the advantages here.

1. It can be done in online. There is a table level lock for very short moment. Traditional methods are not supporting to reset the HWM in online.

2. It does not take extra space while resetting the HWM. If we use traditional method, DBMS_REDEFINITION package use double the amount of space.

3. It does acquire only row level lock while performing majority of the shrinking(moving rows) work. It acquires table level lock only when it resets the HWM which is in phase II. But traditional methods requires table down time for resetting the HWM except using dbms_redefinition package.

4. Index will be maintained and remain usable. But in traditional methods, we need to rebuild the index. Especially when we use ALTER TABLESPACE MOVE command.

5. It can be made in one command(alter table emp shrink space). In traditional method, there are multiple steps.

6. If you are not sure that you can afford table level lock at specific time, then you can do the majority of the shriniking work and later we can reset the HWM. Since table level lock is required only while resetting the HWM. The whole process can be done in two steps as i explained above. This advantage is not available in traditional methods.

What are the restriction of using Oracle10g new feature to reset the HWM?

1. It is only possible in ASSM tablespace
2. Not supporting for clustered tables, tables with column data type LONG

What circumstances we can reset the HWM as two phase?

The table is not frequently used(insert/update/delete) and it can afford to have table level lock, then we can reset the HWM in one go. We can reset the HWM as two steps when the table is used by several people and it is always busy and it does not permit the table level lock even for short moment at specific time. This scenario, we can move the rows to shrink the table. Then in the night time or off peak hours, we can reset the HWM.

Tuesday, September 15, 2009

DBMS_PROFILER

DBMS_PROFILER is introduced in oracle8i. It is powerful tool to find the PLSQL execution time and determine the bottleneck of the program unit. This tool provides information about how many times each line is executing and how much time it is spending to execute for each line of the code. The basic idea behind the profiling is, developers can understand where the code is taking most time and they can detect and optimize the PLSQL code. We use SQL trace to determine the bottleneck for SQL code. But for PLSQL code, we can use dbms_profiler utility to profile the run time behaviour. Steps might very for other oracle versions.

How do we set up dbms_profiler utility?

Please remember, the dbms_profiler setup is not part of Oracle installation. We need to setup manually if we want to profile the PLSQL code. There are five simple steps to configure the dbms_profiler. Let us start configure the profiler. This article is tested in oracle10g R2.

Step1. The dbms_profiler package can be loaded by running the $ORACLE_HOME/rdbms/admin/profload.sql script as the SYS user. Execute profload.sql in sys schema.

Connected to:
Oracle Database 10g Enterprise Edition Release 10.2.0.3.0 - Production
With the Partitioning, OLAP and Data Mining options

SQL> connect sys/password@orcl as sysdba
Connected.
SQL> start c:/oracle/product/10.2.0/db_1/rdbms/admin/profload.sql

Package created.

Grant succeeded.

Synonym created.

Library created.

Package body created.

Testing for correct installation
SYS.DBMS_PROFILER successfully loaded.

PL/SQL procedure successfully completed.

Step2. dbms_profiler package requires some schema objects which should be created in central schema or application schema. Create a new schema called profiler. We can either create a new schema or use existing schema. In this case, i created new schema, named Profiler.

SQL> create user profiler
2 identified by profiler;

User created.

SQL> grant create session,resource,connect to profiler;
Grant succeeded.

SQL>

Step3. Run the $ORACLE_HOME/rdbms/admin/proftab.sql file on the profiler schema to create some schema objects to store profiler information. This proftab.sql file creates below three tables with some other objects.

1.PLSQL_PROFILER_RUNS
2.PLSQL_PROFILER_UNITS
3.PLSQL_PROFILER_DATA

SQL> connect profiler/profiler@orcl
Connected.
SQL> start c:/oracle/product/10.2.0/db_1/rdbms/admin/proftab.sql
drop table plsql_profiler_data cascade constraints
*
ERROR at line 1:
ORA-00942: table or view does not exist

drop table plsql_profiler_units cascade constraints
*
ERROR at line 1:
ORA-00942: table or view does not exist

drop table plsql_profiler_runs cascade constraints
*
ERROR at line 1:
ORA-00942: table or view does not exist

drop sequence plsql_profiler_runnumber
*
ERROR at line 1:
ORA-02289: sequence does not exist

Table created.

Comment created.

Table created.

Comment created.

Table created.

Comment created.

Sequence created.

SQL>

Step4. Connect into profiler schema and grant the below privileges.

GRANT SELECT ON plsql_profiler_runnumber TO PUBLIC;
GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_data TO PUBLIC;
GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_units TO PUBLIC;
GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_runs TO PUBLIC;

SQL> connect profiler/profiler@orcl
Connected.
SQL> GRANT SELECT ON plsql_profiler_runnumber TO PUBLIC;

Grant succeeded.

SQL> GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_data TO PUBLIC;

Grant succeeded.

SQL> GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_units TO PUBLIC;

Grant succeeded.

SQL> GRANT SELECT, INSERT, UPDATE, DELETE ON plsql_profiler_runs TO PUBLIC;

Grant succeeded.

SQL>

Step5. Connect into sys schema and grant the below privileges for dbms_profiler package.

CREATE PUBLIC SYNONYM plsql_profiler_runs FOR profiler.plsql_profiler_runs;
CREATE PUBLIC SYNONYM plsql_profiler_units FOR profiler.plsql_profiler_units;
CREATE PUBLIC SYNONYM plsql_profiler_data FOR profiler.plsql_profiler_data;
CREATE PUBLIC SYNONYM plsql_profiler_runnumber FOR profiler.plsql_profiler_runnumber;

SQL> connect sys/password@orcl as sysdba
Connected.
SQL> CREATE PUBLIC SYNONYM plsql_profiler_runs FOR profiler.plsql_profiler_runs;

Synonym created.

SQL> CREATE PUBLIC SYNONYM plsql_profiler_units FOR profiler.plsql_profiler_units;

Synonym created.

SQL> CREATE PUBLIC SYNONYM plsql_profiler_data FOR profiler.plsql_profiler_data;

Synonym created.

SQL> CREATE PUBLIC SYNONYM plsql_profiler_runnumber FOR profiler.plsql_profiler_runnumber;

Synonym created.

SQL>

Once we are successful with five steps, we can start profiling the PLSQL code.

How do we profile the PLSQL Procedure?

Let us create sample procedure and profile the procedure.

1. start profiler
2. run the procedure
3. stop profiler
4. flush data from memory and save into table
5. Analyze the data and see where it is taking more time

Here i create a procedure called do_something in SCOTT schema. You can also profile the procedure which is existing in any schema in the database.

SQL> CREATE OR REPLACE PROCEDURE do_something (p_times IN NUMBER) AS
2 v_cnt NUMBER;
3 BEGIN
4 FOR i IN 1 .. p_times LOOP
5 SELECT count(*) + p_times
6 INTO v_cnt
7 FROM EMP;
8 END LOOP;
9 END;
10 /

Procedure created.

The below unnamed PLSQL code starts the profiler and call the procedure. Once the procedure is executed, it stop the profiler. It flush the data from memory and save into table.

SQL> DECLARE
2 l_result BINARY_INTEGER;
3 BEGIN
4 l_result := DBMS_PROFILER.start_profiler(run_comment => 'do_something: ' SYSDATE);
5 do_something(p_times => 100);
6 l_result := DBMS_PROFILER.stop_profiler;
7 dbms_profiler.flush_data;
8 END;
9 /

PL/SQL procedure successfully completed.

SQL>

Here is query to check the profiler result.

SQL> SET LINESIZE 200
SQL> SET TRIMOUT ON
SQL>
SQL> COLUMN runid FORMAT 99999
SQL> COLUMN run_comment FORMAT A50
SQL> SELECT runid,
2 run_date,
3 run_comment,
4 run_total_time
5 FROM plsql_profiler_runs
6 ORDER BY runid;

RUNID RUN_DATE RUN_COMMENT RUN_TOTAL_TIME
------ --------- -------------------------------------------------- --------------
4 15-SEP-09 do_something: 15-SEP-09 686370753

SQL> SELECT d.line#,
2 d.total_occur,
3 d.total_time
4 FROM plsql_profiler_units u
5 JOIN plsql_profiler_data d ON u.runid = d.runid AND u.unit_number = d.unit_number
6 WHERE u.runid = 4
7 and unit_name='DO_SOMETHING'
8 and unit_owner='SCOTT'
9 and unit_type='PROCEDURE'
10 ORDER BY u.unit_number, d.line#;

LINE# TOTAL_OCCUR TOTAL_TIME
---------- ----------- ----------
1 1 199466
4 101 2247771
5 100 513261322
9 1 85485

SQL> SELECT line ' : ' text
2 FROM all_source
3 WHERE owner = 'SCOTT'
4 AND type = 'PROCEDURE'
5 AND name = 'DO_SOMETHING';

LINE':'TEXT
----------------------------------------------------------------------------------------------------
1 : PROCEDURE do_something (p_times IN NUMBER) AS
2 : v_cnt NUMBER;
3 : BEGIN
4 : FOR i IN 1 .. p_times LOOP
5 : SELECT count(*) + p_times
6 : INTO v_cnt
7 : FROM EMP;
8 : END LOOP;
9 : END;

9 rows selected.

SQL>

Conclusion : The line number 4 runs 101 times and line number 5 runs 100 times. The procedure spends most of the time at line number 5. Now we figured out exactly where it is taking longer time. We can focus on tuning the line 5 in case if we want to....

Where do we use dbms_profiler? TKPROF and Explain plan helps to find where it is taking long time to complete the SQL. Dbms_profiler is useful, if we want to find out which line is consuming most time in entire PLSQL.

The source of this article is oracle-base.

Thursday, September 10, 2009

Oracle SQL Questions for Newbies

I thought, it would be useful if i post some sample SQL questions for newbies. These bunch of SQL questions would be helpful for beginners who wanted to learn Oracle SQL.

Here are some tables which needs to be created to practice SQL questions. Just for learning process, you could use SCOTT schema to practice SQL questions. I am posting 101 Oracle SQL questions with answers.

First step would be, we need to create the below tables and insert relevant data to practice the questions.

Create the below tables:

CREATE TABLE DEPT (
DEPTNO NUMBER(2),
DNAME VARCHAR2(14),
LOC VARCHAR2(13));

INSERT INTO DEPT VALUES (10, 'ACCOUNTING', 'NEW YORK');
INSERT INTO DEPT VALUES (20, 'RESEARCH', 'DALLAS');
INSERT INTO DEPT VALUES (30, 'SALES', 'CHICAGO');
INSERT INTO DEPT VALUES (40, 'OPERATIONS', 'BOSTON');

ALTER TABLE DEPT ADD PRIMARY KEY(DEPTNO);

CREATE TABLE EMP (EMPNO NUMBER(4) NOT NULL,
ENAME VARCHAR2(10),
JOB VARCHAR2(9),
MGR NUMBER(4),
HIREDATE DATE,
SAL NUMBER(7,2),
COMM NUMBER(7,2),
DEPTNO NUMBER(2));

ALTER TABLE EMP ADD CONSTRAINT FK_EMP_01
FOREIGN KEY(DEPTNO) REFERENCES DEPT;

INSERT INTO EMP VALUES (7369, 'SMITH', 'CLERK', 7902, TO_DATE('17-DEC-1980', 'DD-MON-YYYY'), 800, NULL, 20);
INSERT INTO EMP VALUES (7499, 'ALLEN', 'SALESMAN', 7698, TO_DATE('20-FEB-1981', 'DD-MON-YYYY'), 1600, 300, 30);
INSERT INTO EMP VALUES (7521, 'WARD', 'SALESMAN', 7698, TO_DATE('22-FEB-1981', 'DD-MON-YYYY'), 1250, 500, 30);
INSERT INTO EMP VALUES (7566, 'JONES', 'MANAGER', 7839, TO_DATE('2-APR-1981', 'DD-MON-YYYY'), 2975, NULL, 20);
INSERT INTO EMP VALUES (7654, 'MARTIN', 'SALESMAN', 7698,TO_DATE('28-SEP-1981', 'DD-MON-YYYY'), 1250, 1400, 30);
INSERT INTO EMP VALUES (7698, 'BLAKE', 'MANAGER', 7839,TO_DATE('1-MAY-1981', 'DD-MON-YYYY'), 2850, NULL, 30);
INSERT INTO EMP VALUES (7782, 'CLARK', 'MANAGER', 7839,TO_DATE('9-JUN-1981', 'DD-MON-YYYY'), 2450, NULL, 10);
INSERT INTO EMP VALUES (7788, 'SCOTT', 'ANALYST', 7566,TO_DATE('09-DEC-1982', 'DD-MON-YYYY'), 3000, NULL, 20);
INSERT INTO EMP VALUES (7839, 'KING', 'PRESIDENT', NULL,TO_DATE('17-NOV-1981', 'DD-MON-YYYY'), 5000, NULL, 10);
INSERT INTO EMP VALUES (7844, 'TURNER', 'SALESMAN', 7698,TO_DATE('8-SEP-1981', 'DD-MON-YYYY'), 1500, 0, 30);
INSERT INTO EMP VALUES (7876, 'ADAMS', 'CLERK', 7788,TO_DATE('12-JAN-1983', 'DD-MON-YYYY'), 1100, NULL, 20);
INSERT INTO EMP VALUES (7900, 'JAMES', 'CLERK', 7698,TO_DATE('3-DEC-1981', 'DD-MON-YYYY'), 950, NULL, 30);
INSERT INTO EMP VALUES (7902, 'FORD', 'ANALYST', 7566,TO_DATE('3-DEC-1981', 'DD-MON-YYYY'), 3000, NULL, 20);
INSERT INTO EMP VALUES (7934, 'MILLER', 'CLERK', 7782,TO_DATE('23-JAN-1982', 'DD-MON-YYYY'), 1300, NULL, 10);

CREATE TABLE SALGRADE(
GRADE NUMBER(2),
LOSAL NUMBER,
HISAL NUMBER);

INSERT INTO SALGRADE VALUES(1, 700,1200);
INSERT INTO SALGRADE VALUES(2, 1201,1400);
INSERT INTO SALGRADE VALUES(3, 1401,2000);
INSERT INTO SALGRADE VALUES(4, 2001,3000);
INSERT INTO SALGRADE VALUES(5, 3001,9999);

Question & Answers:

1) Display all the records in emp table?
select * from emp;
2) Display all the records in emp table where employee belongs to deptno 10?
select * from emp where deptno = 10

3) Display all the records in emp table where employee does not belong to deptno 30?

select * from emp where deptno != 30;

4) Display total number of records in Emp table?

select count(*) from emp;

5) Display emp table with salary descending order?

select * from emp order by sal desc

6) Display first five records in employee table?

select * from emp where rownum <= 5
7) Display all the records in emp table order by ascending deptno, descending salary?
select * from emp order by deptno asc, sal desc

8) Display all employees those who were joined in year 1981?

select * from emp where to_char(hiredate,'YYYY') = 1981;

9) Display COMM in emp table. Display zero in place of null.

select nvl(comm,0) from emp

10) Display the records in emp table where MGR in 7698,7566 and sal should be greater then 1500

select * from emp where mgr in(7698,7566) and sal > 1500

11) Display all employees where employees hired before 01-JAN-1981

select * from emp where hiredate < '01-JAN-1981'
12) Display all employees with how many years they have been servicing in the company?

select hiredate,round((sysdate-hiredate)/360) as years from emp

13) Display all employees those were not joined in 1981?

select * from emp where to_char(hiredate,'YYYY') != 1981;

14) Display all employees where their hiredate belongs to third quarter?

select * from emp where to_char(hiredate,'Q') = 3;

15) Display all employees where their salary is less then the Ford’s salary?

select * from emp where sal <(select sal from emp where ename='FORD');
16) Display all the records in EMP table along with the rowid?

select ename,rowid from emp;

17) Display all records in EMP table those were joined before SCOTT joined?

select * from emp where hiredate <(select hiredate from emp where ename='SCOTT')
18) Display all employees those who were joined in third quarter of 1981?

select * from emp where to_char(hiredate,'Q') = 3 and to_char(hiredate,'YYYY') = 1981

19) Add 3 months with hiredate in EMP table and display the result?

select hiredate, add_months(hiredate,3) from emp

20) Display the date for next TUESDAY in hiredate column?

select next_day(hiredate,'TUESDAY') from emp;

21) Find the date, 15 days after today’s date.

select sysdate+15 from dual

22) Write a query to display current date?

select sysdate from dual;
select current_date from dual;

23) Display distinct job from emp table?

select distinct job from emp

24) Display all the records in emp table where employee hired after 28-SEP-81 and before 03-DEC-81?

select * from emp where hiredate between '28-SEP-81' and '03-DEC-81'

25) Write a query that displays the employee’s names with the first letter capitalized and all other letters lowercase for all employees whose name starts with J, A, or M

select initcap(ename) from emp where ename like 'J%' or ename like 'A%' or ename like 'M%'

26) Display all jobs that are in department 10. Include the location of department in the output.

select job, loc from emp,dept where emp.deptno = dept.deptno and emp.deptno =10

27) Write a query to display the employee name, department name of all employees who earn a commission

select ename,dname from emp,dept where emp.deptno = dept.deptno and comm is not null;

28) Display the empno, ename, sal, and salary increased by 15%.

select empno, ename, sal actual_sal, (sal * 15/100) as Increased_sal from emp

29) Display ename, sal, grade. Use emp, salgrade table

select ename,sal,grade from emp,salgrade where sal between losal and hisal;

30) Display all employees and corresponding managers

select w.ename,w.sal,m.ename,m.sal from emp w, emp m where w.mgr = m.empno;

31) Display all the departments where employee salary greater then average salary of that department.

select ename,deptno, sal from emp a where sal > (select avg(sal) from emp where emp.deptno = a.deptno) order by deptno;

32) Display all employees whose salary greater then the manager salary?

select w.ename,w.sal,m.ename,m.sal from emp w, emp m where w.mgr = m.empno and w.sal > m.sal

33) Display employees where length of ename is 5

select * from emp where length(ename) =5

34) Display all employees where ename start with J and ends with S

select * from emp where ename like 'J%S'

35) Display all employees where employee does not belong to 10,20,40

select * from emp where deptno not in(10,20,40)

36) Display all employees where jobs does not belong to PRESIDENT and MANAGER?

select * from emp where job not in('PRESIDENT','MANAGER');

37) Display the maximum salary in the emp table

select max(sal) from emp

38) Display average salary for job SALESMAN

select avg(sal) from emp where job = 'SALESMAN'

39) Display all three figures salary in emp table

select * from emp where sal < = 999;
select * from emp where length(sal) = 3;
40) Display all records in emp table for employee who does not receive any commission

select * from emp where comm is not null

41) Display all ename where first character could be anything, but second character should be L?
select * from emp where ename like '_L%'

42) Display nth highest and nth lowest salary in emp table?

SELECT DISTINCT (a.sal) FROM EMP A WHERE &N = (SELECT COUNT (DISTINCT (b.sal)) FROM EMP B WHERE a.sal<=b.sal);
select distinct sal from (select ename,sal,dense_rank() over(order by sal desc) dr from emp) where dr = &x ;
43) Display all the departments where department has 3 employees?

select deptno from dept a where deptno in(select deptno from emp group by deptno having count(*)=3)

44) Display emp name and corresponding subordinates. Use CONNECT BY clause.

select lpad(' ',level+12)+ename from emp connect by prior empno = mgr start with mgr is null

Note: Please replace pipe symbol in the place of + sign for question 44. Pipe symbol is not displaying the blog. This is the reason, i used Plus sign here.

45) Display sum of salary for each department. The output should be in one record

select sum(decode(deptno,10,sal)) dept10, sum(decode(deptno,20,sal)) dept20, sum(decode(deptno,30,sal)) dept30, sum(sal) total_sal from emp

46) Display all department with Minimum salary and maximum salary?

select min(sal),max(sal) from emp;

47) Display all ename, sal, deptno,dname from emp, dept table where all department which has employees as well as department does not have any employees. This query should include non matching rows.

select dname,b.deptno, ename,sal from emp a, dept b where a.deptno(+) = b.deptno;
select dname,b.deptno, ename,sal from emp a right outer join dept b on a.deptno = b.deptno;

48) Display all ename, sal, deptno from emp, dept table where all employees which has matching department as well as employee does not have any departments. This query should include non matching rows.
Note: In the below query, employee will always have matching record in dept table. Emp, dept table may not be good example to answer this question.
select dname,b.deptno, ename,sal from emp a, dept b where a.deptno = b.deptno(+);
select dname,b.deptno, ename,sal from emp a left outer join dept b on a.deptno = b.deptno;
49) Display all ename, sal, deptno from emp, dept table where all employees which has matching and non matching department as well as all departments in dept table which has matching and non matching employees. This query should include non matching rows on both the tables.
Note: In the below query, employee will always have matching record in dept table. Emp, dept table may not be good example to answer this question.
select dname,b.deptno, ename,sal from emp a full outer join dept b on a.deptno = b.deptno
50) Display all ename, empno, dname, loc from emp, dept table without joining two tables
select * from emp,dept;

51) Display all the departments where department does not have any employees

select deptno from dept where not exists(select 1 from emp where emp.deptno = dept.deptno);

select deptno from dept where deptno not in(select deptno from emp);

52) Display all the departments where department does have atleast one employee

select * from dept a where exists(select 1 from emp b where b.deptno = a.deptno)

select * from dept a where deptno in(select deptno from emp b where a.deptno = b.deptno)

53) Display all employees those who are not managers?

select ename from emp a where not exists (select 1 from emp b where b.mgr = a.empno);

select ename from emp a where empno not in (select mgr from emp b where b.mgr = a.empno and mgr is not null);

54) Display ename, deptno from emp table with format of {ename} belongs to {deptno}

select ename+' belongs to '+deptno from emp

Note: Please replace pipe symbol in the place of + sign for question 44. Pipe symbol is not displaying the blog. This is the reason, i used Plus sign here.

55) Display total number of employees hired for 1980,1981,1982. The output should be in one record.

select
count(decode(to_char(hiredate,'YYYY'), 1980,hiredate)) total_hire_1980,
count(decode(to_char(hiredate,'YYYY'), 1981,hiredate)) total_hire_1981,
count(decode(to_char(hiredate,'YYYY'), 1982,hiredate)) total_hire_1982
from emp

56) Display ename, deptno from employee table. Also add another column in the same query and it should display ten for dept 10, twenty for dept 20, thirty for dept 30, fourty for dept 40

select ename,deptno, (case deptno
when 10 then 'Ten'
when 20 then 'Twenty'
when 30 then 'Thirty'
when 40 then 'fourty'
else 'others' end) as dept
from emp

57) Display all the records in emp table. The ename should be lower case. The job first character should be upper case and rest of the character in job field should be lower case.

select lower(ename) as ename, initcap(job) as job from emp
58) Display all employees those who have joined in first week of the month ?

select * from emp where to_char(hiredate,'W') = 1;

59) Display all empoyees those who have joined in the 49th week of the year?

select * from emp where to_char(hiredate,'WW') = 49;

60) Display empno, deptno, salary, salary difference between current record and previous record in emp table. Deptno should be in descending order.

SELECT empno,
ename,
job,
sal,
LAG(sal, 1, 0) OVER (ORDER BY sal) AS sal_prev,
sal - LAG(sal, 1, 0) OVER (ORDER BY sal) AS sal_diff
FROM emp;

61) Create table emp1 and copy the emp table for deptno 10 while creating the table

Create table emp1 as select * from emp where deptno=10

62) Create table emp2 with same structure of emp table. Do not copy the data

create table emp2 as select * from emp where 1=2

63) Insert new record in emp1 table, Merge the emp1 table on emp table.

insert into emp1 values(9999,'PAUL','MANAGER',7839,SYSDATE,8900,NULL,10);

MERGE
INTO emp tgt
USING emp1 src
ON ( src.empno = tgt.empno )
WHEN MATCHED
THEN
UPDATE
SET tgt.ename = src.ename,
tgt.job = src.job,
tgt.mgr = src.mgr,
tgt.hiredate = src.hiredate,
tgt.sal = src.sal,
tgt.deptno = src.deptno
WHEN NOT MATCHED
THEN
Insert(
Tgt.empno,
Tgt.Ename,
Tgt.Job,
Tgt.Mgr,
Tgt.Hiredate,
Tgt.Sal,
Tgt.Comm,
Tgt.Deptno)
values (src.empno,
src.ename,
src.job,
src.mgr,
src.hiredate,
src.sal,
src.comm,
src.deptno);

64) Display all the records for deptno which belongs to employee name JAMES?

select * from emp where deptno in(select deptno from emp where ename = 'JAMES')

65) Display all the records in emp table where salary should be less then or equal to ADAMS salary?

select * from emp where sal <= (select sal from emp where ename='ADAMS')

66) Display all employees those were joined before employee WARD joined?

select * from emp where hiredate < (select hiredate from emp where ename='WARD')

67) Display all subordinate those who are working under BLAKE?

Select ename from emp where mgr = (select empno from emp where ename='BLAKE')

68) Display all subordinate(all levels) for employee BLAKE?

select ename from emp start with empno = (select empno from emp where ename='BLAKE')
connect by prior empno = mgr

69) Display all record in emp table for deptno which belongs to KING's Job?

select * from emp where deptno in(select deptno from emp where job= (select job from emp where ename = 'KING'))

70) Display the employees for empno which belongs to job PRESIDENT?

select * from emp where empno in(select empno from emp where ename in(select ename from emp where JOB = 'PRESIDENT'));

71) Display list of ename those who have joined in Year 81 as MANAGER?

select * from emp where to_char(hiredate,'YYYY') = 1981 and job = 'MANAGER';

72) Display who is making highest commission?

select * from emp where comm = (select max(comm) from emp);

73) Display who is senior most employee? How many years has been working?

select * from emp where trunc(sysdate-hiredate)/365 = (select max(trunc(sysdate-hiredate)/365) from emp);

select * from emp where hiredate =(select min(hiredate) from emp)
74) Display who is most experienced and least experienced employee?

select * from emp where trunc(sysdate-hiredate)/365 = (select min(trunc(sysdate-hiredate)/365) from emp);

select * from emp where hiredate =(select max(hiredate) from emp)

75) Display ename, sal, grade, dname, loc for each employee.

select empno,ename,b.deptno,dname,grade from
emp a,dept b, salgrade c
where a.deptno = b.deptno
and sal between losal and hisal;

76) Display all employee whose location is DALLAS?

SELECT emp.ename, emp.JOB, emp.deptno
FROM emp
WHERE EXISTS
(SELECT 'x'
FROM dept d
WHERE d.DEPTNO = emp.DEPTNO
AND d.LOC = 'DALLAS') ;

select emp.ename, emp.job, emp.deptno
from emp
where deptno in(select deptno from dept where loc='DALLAS');

77) Display ename, job, dname, deptno for each employee by using INLINE view?

SELECT emp.ename,
emp.JOB,
emp.deptno,
dnames.dname
FROM emp
JOIN (select dname, deptno
from dept ) dnames ON emp.deptno = dnames.deptno

78) List ename, job, sal and department of all employees whose salary is not within the salary grade?

select ename, job, sal, dname
from emp, dept
where emp.deptno = dept.deptno
and not exists
(select ‘x’ from salgrade
where emp.sal between losal and hisal);

79) Use EMP and EMP1 table. Query should have only three columns. Display empno,ename,sal from both tables inluding duplicates.
select empno, ename, sal from emp
union all
select empno, ename, sal from emp1

80) Delete emp table for detpno 10 and 20.
delete emp where deptno in(10,20);

81) Delete all employees those are not getting any commission?
delete emp where comm is null;

82) Delete all employees those who employeed more then 28 years
delete emp where trunc(sysdate - hiredate)/365 > 28;

83) Add duplicate records in emp1 table. Delete the duplicate records in emp1 table.
insert into emp1 select * from emp1 where rownum <=1; commit; delete emp1 a where a.rowid <>(select min(b.rowid) from emp1 b where a.empno = b.empno);

84) Delete the employees where employee salary greater then average salary of department salary?
delete emp a where sal > (select avg(sal) from emp where emp.deptno = a.deptno);

85) Delete all employees those who are reporting to BLAKE?
Delete emp where ename in(Select ename from emp where mgr = (select empno from emp where ename='BLAKE'))

86) Delete all levels of employees those who are under BLAKE?
Delete emp where ename in(select ename from emp start with empno = (select empno from emp where ename='BLAKE')
connect by prior empno = mgr)

87) Delete all employees those who are only managers?
delete emp where ename in(select ename from emp a where empno in (select mgr from emp b where b.mgr = a.empno and mgr is not null))

88) Remove the department in dept table where dept does not have any employees?

delete dept where deptno not in(select deptno from emp where deptno is not null)

89) Remove all grade 2 employees in emp table?

delete emp where empno in(select empno from emp,salgrade where sal between losal and hisal and grade = 2)

90) Remove all the employees in SMITH's department

delete emp where deptno = (select deptno from emp where ename = 'SMITH')

91) Remove least paid employee who are reporting to BLAKE ?

delete emp where sal = (select min(sal) from emp where mgr =
(select empno from emp where ename = 'BLAKE')) and
ename in(select ename from emp where mgr =
(select empno from emp where ename = 'BLAKE'))

92) Remove all employees who were joined before SMITH joined?
delete emp where hiredate < (select hiredate from emp where ename='SMITH');
93) Rename the employee name JONES to ANDY


update emp set ename = 'ANDY' where ename = 'JONES'

94) Change the WARD's hiredate to one day ahead

update emp set hiredate = hiredate + 1 where ename = 'WARD'

95) Update MARTIN salary same as SMITH's salary

update emp set sal = (select sal from emp where ename = 'SMITH') where ename='MARTIN'

96) Increase the salary 5% for employee those who are earning commission less then 1000

update emp set sal = sal + (sal * (5/100)) where comm between 0 and 1000

97) Increase 250$ commission for BLAKE's team

update emp set comm = nvl(comm,0)+250 where mgr = (select empno from emp where ename='BLAKE');
98) Increase 100$ for employee who is making more then averge salary of his department?

update emp a set sal = sal + 150 where sal > (select avg(sal) from emp b where b.deptno = a.deptno)
99) Increase 1% salary for employee who is making lowest salary in dept 10

update emp set sal = sal + (sal* 1/100)
where
sal = (select min(sal) from emp where deptno = 10)
and deptno = 10

100) Reduce the commission amount from employee salary for each employee who were joined after ALLEN joined.

update emp set sal = sal - NVL(comm,0)
where empno in(select empno from emp where hiredate > (select hiredate from emp where
ename = 'ALLEN'))

101) Increase commission 10$ for employees those who are located in NEW YORK.

update emp a set comm = NVL(COMM,0) + 10
where deptno = (select deptno from dept where loc='NEW YORK');

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