问题
To find out the Nth max sal in oracle i'm using below query
SELECT DISTINCE sal
FROM emp a
WHERE (
SELECT COUNT(DISTINCE sal)
FROM emp b
WHERE a.sal<=b.sal)=&n;
But According to me by using the above query it will take more time to execute if table size is big.
i'm trying to use the below query
SELECT sal FROM ( SELECT DISTINCE sal FROM emp ORDER BY sal DESC ) WHERE rownum=3;but not getting output.. any suggetions please .. Please share any link on how to optimise queries and decrease the time for a query to execute.
回答1:
try this
select *
from
(
select
sal
,dense_rank() over (order by sal desc) ranking
from table
)
where ranking = 4 -- Replace 4 with any value of N
回答2:
SELECT sal FROM (
SELECT sal, row_number() OVER (order by sal desc) AS rn FROM emp
)
WHERE rn = 3
Yes, it will take longer to execute if the table is big. But for "N-th row" queries the only way is to look through all the data and sort it. It will be definitely much faster if you have an index on sal.
回答3:
SELECT *
FROM Employee Emp1
WHERE (N-1) = (
SELECT COUNT(DISTINCT(Emp2.Salary))
FROM Employee Emp2
WHERE Emp2.Salary > Emp1.Salary)
回答4:
We could write as below mentioned also.
select min(sal) from (select sal from emp where rownum=<&n order by sal desc);
回答5:
This will show the 3rd max salary from table employee.
If you want to find out the 5th or 6th (whatever you want) value then just change the where condition like this where rownum<=5" or "where rownum<=6 and so on...
select min(sal) from(select distinct(sal) from emp where rownum<=3 order by sal desc);
回答6:
SELECT sal
FROM (
SELECT empno,
deptno, sal,
dense_rank( ) over ( partition by deptno order by sal desc) NRANK
FROM emp
)
WHERE NRANK = 4
回答7:
SELECT *
FROM (
SELECT empno,
deptno, sal,
dense_rank( ) over ( order by sal desc) NRANK
FROM emp
)
WHERE NRANK = 4
回答8:
you can replace the 2 with your desired number
select * from ( select distinct (sal),ROW_NUMBER() OVER (order by sal desc) rn from emp ) where rn=2
回答9:
Refer following query for getting nth highest salary. By this way you get nth highest salary. If you want get nth lowest salary only you need to replace DESC by ASC in the query.
回答10:
Now you try this you will get for sure:
SELECT DISTINCT sal
FROM emp a
WHERE (
SELECT COUNT(DISTINCT sal)
FROM emp b
WHERE a.sal<=b.sal)=&n;
For your information, if you want the nth least sal:
SELECT DISTINCT sal
FROM emp a
WHERE (
SELECT COUNT(DISTINCT sal)
FROM emp b
WHERE a.sal>=b.sal)=&n;
回答11:
select min(sal) from (select distinct sal from employee order by sal DESC) where rownum<=N;
place the number whatever the highest sal you want to retrieve.
回答12:
Try out following:
SELECT *
FROM
(SELECT rownum AS rn,
a.*
FROM
(WITH DATA AS -- creating dummy data
( SELECT 'MOHAN' AS NAME, 200 AS SALARY FROM DUAL
UNION ALL
SELECT 'AKSHAY' AS NAME, 500 AS SALARY FROM DUAL
UNION ALL
SELECT 'HARI' AS NAME, 300 AS SALARY FROM DUAL
UNION ALL
SELECT 'RAM' AS NAME, 400 AS SALARY FROM DUAL
)
SELECT D.* FROM DATA D ORDER BY SALARY DESC
) A
)
WHERE rn = 3; -- specify N'th highest here (In this case fetching 3'rd highest)
Cheers!
回答13:
select * FROM (
select EmployeeID, Salary
, dense_rank() over (order by Salary DESC) ranking
from Employee
)
WHERE ranking = N;
dense_rank() is used for the salary has to be same.So it give the proper output instead of using rank().
回答14:
SELECT TOP (1) Salary FROM
(
SELECT DISTINCT TOP (10) Salary FROM Employee ORDER BY Salary DESC
) AS Emp ORDER BY Salary
This is for 10th max salary, you can replace 10 with n.
回答15:
These queries will also work:
Workaround 1)
SELECT ename, sal
FROM Emp e1 WHERE n-1 = (SELECT COUNT(DISTINCT sal)
FROM Emp e2 WHERE e2.sal > e1.sal)
Workaround 2) using row_num function.
SELECT *
FROM (
SELECT e.*, ROW_NUMBER() OVER (ORDER BY sal DESC) rn FROM Emp e
) WHERE rn = n;
Workaround 3 ) using rownum pseudocolumn
Select MAX(SAL)
from (
Select *
from (
Select *
from EMP
order by SAL Desc
) where rownum <= n
)
回答16:
This will also work :
with data as
(
select sal,rwid from (
select salary as sal,rowid as rwid from salary order by salary desc
)
where rownum < 5
)
select * from salary a
where rowid = (select min(rwid) from data)
回答17:
select min(sal) from (select distinct(sal) from emp order by sal desc) where rownum <=&n;
Inner query select distinct(sal) from emp order by sal desc will give the below output as given below.
SAL 5000 3000 2975 2850 2450 1600 1500 1300 1250 1100 950 800
without distinct in the above query select sal from emp order by sal desc
output as given below.
SAL 5000 3000 3000 2975 2850 2450 1600 1500 1300 1250 1250 1100 950 800
outer query will give the 'N'th max sal (E.g) I have tried here for 4th Max sal and out put as given below.
MIN(SAL) 2850
回答18:
Select min(salary) from (
select distinct(salary) from empdetails order by salary desc
) where rownum <=&rn
Just enter nth number which you want.
回答19:
Try this:
SELECT min(sal) FROM (
SELECT sal FROM emp ORDER BY sal desc) WHERE ROWNUM <= 3; -- Replace 3 with any value of N
回答20:
You can optimize the query using Dense_rank() function.
for Example :
select distinct salary from ( select salary ,dense_rank() over (order by salary desc) ranking from Employee ) where ranking = 6
Note: ranking 6 is the number of nth order.
回答21:
SELECT Min(sal)
FROM (SELECT DISTINCT sal
FROM emp
WHERE sal IS NOT NULL
ORDER BY sal DESC)
WHERE rownum <= n;
回答22:
5th highest salary:
SELECT
*
FROM
emp a
WHERE
4 = (
SELECT
COUNT(DISTINCT b.sal)
FROM
emp b
WHERE
a.sal < b.sal
)
Replace 4 with any value of N.
回答23:
There are three methods are there...
SELECT salary,first_name,rnk
FROM (SELECT salary,first_name,rank() over (order by salary desc nulls last) as rnk from emp) where rnk=3;
SELECT salary,first_name,rnk
FROM (SELECT salary,first_name,dense_rank() over (order by salary desc nulls last) as rnk from emp) where rnk=3;
select rnk,first_name,salary
from (select rownum as rnk ,first_name,salary
from (select first_name,salary
from emp order by salary desc nulls last)) where rnk=3
来源：https://stackoverflow.com/questions/20091125/nth-max-salary-in-oracle