+ 36
How to find second highest salary of a given table in oracle database
16 ответов
+ 15
SELECT max(salary) FROM Employee WHEREsalary < (SELECT max(salary) FROM Employee);
+ 30
Sorry sorry it's execute thank u so much Adarsh Srivastava 🙏🙏🙏
+ 22
Adarsh Srivastava only second highest salary..
If we used this command then all rows selected expected highest salary of a table..
+ 8
Mind that this is an untested query, an alternative by using ROWNUM limiter:
SELECT salary FROM payroll WHERE ROWNUM = 2 ORDER BY salary DESC;
+ 8
Ipang If this doesn't work, the ORDER BY clause may need to be applied as a subquery.
Example:
SELECT * FROM
( SELECT salary
FROM payroll
ORDER BY salary DESC )
WHERE ROWNUM = 2;
+ 6
[aside] Is this homework? I ask because I see this question semi-regularly with minor changes that don't make the problem that's actually being solved unique.
One top-level abstraction could cover this case and numerous variations (vs the one-off's), but the value of an abstraction depends on whether this is likely to keep repeating.
+ 6
Yes David Carroll that probably how I should've written it in the first place : )
I can't test it though, nowhere near any server, but your query looks more selective than my proposal.
Thanks David ; )
+ 4
Jyoti I dont think so....It will give 2nd highest salary surely..
+ 3
Jyoti no need to be sorry...My pleasure
+ 3
Not a comment on the challenge or in a working technique, just on how I write some queries.
I have good success using Common Table Expressions (CTE). It's a formalized technique to combine more than one query as named entities.
It's usually pretty fast for me to come up with good answers to complex queries.
Opening word is WITH:
WITH name1 AS ( query_1 ), name2 AS ( query_2 ) query_3 ;
If there are more queries to add, the process is repreated with all subqueries, and followed by the outside query.
WITH COSAL AS
( SELECT salary , empid, fname, lname, dept
FROM payroll
ORDER BY salary DESC
)
SELECT salary
FROM COSAL
WHERE ROWNUM = 2
;
I just find this format easier to read, construct, and use. If I need to design as I go, this is the simplest way for me to keep things straight.
+ 3
Use ranking
select *
from ( select salary
, dense_rank() over (order by salary desc) ranking
from employee )
where ranking = 2 -- Replace 2 with any value of N
+ 1
How do you mean pls
+ 1
Select salary from (Select salary from Employee order by salary desc limit 1) order by salary aesc limit 1
+ 1
Select max(salary) from Employee where salary <>(Select max(salary)from Employee)
0
Тут есть русский
0
Nice joti