“How can I select the Nth highest salary of the EMP table?”
This is a question that every Oracle newbie stumbles over. Ask it on a forum and you’re pointed to the archives. That gets you nowhere as when you search the archives, all you find is a host of other messages also asking you to search the archives.
Here comes the answer to the problem of finding the Nth highest salary. You can extend this logic to find the Nth highest row of any table.
First things first: The question is ambiguous!
Let’s say this is your data:
Name  Salary 
KING  5000 
FORD  3000 
SCOTT  3000 
JONES  2975 
BLAKE  2850 
CLARK  2850 
ALLEN  1600 
Who is second – FORD or SCOTT or both?
What will you say about JONES’s salary – is it the 3rd highest salary, or the 4th highest?
If you are looking for the set of people earning the Nth highest salary, with no gaps in case of ties, then JONES should be ranked 3rd, after KING [5000, 1st], followed by FORD and SCOTT [both 3000, 2nd].
If you are looking for exact ranks with gaps if there are ties, then JONES is the 4th highest paid employee, as there are 3 people earning more than him – KING, FORD and SCOTT. In this system of ranking, FORD and SCOTT are 2nd jointly and no employee is 3rd.
This is how your ranks will look, in the 2 cases:

Scenario 1: No gaps in case of ties Scenario 2: Gaps in case of ties
Once you have your question sorted out –
(a) Set of people earning the Nth highest salary, with continuous ranks if there are ties, OR
(b) Set of people earning the Nth highest salary, with skipped rank numbers if there are ties
Then you can proceed to writing the queries.
Scenario 1: DENSE_RANK () for Nth highest row, no gaps in case of ties
The analytic function dense_rank() will rank the rows with no gaps in ranking sequence if there are ties.
The ranks are calculated as:
SQL> select ename 2 ,sal 3 ,dense_rank() over (order by sal desc) ranking 4 from emp; ENAME SAL RANKING    KING 5000 1 FORD 3000 2 SCOTT 3000 2 JONES 2975 3 CLARK 2850 4 BLAKE 2850 4 ALLEN 1600 5
Wrap a filter around and pick out the Nth highest salary, say the 4th highest salary.
SQL> select * 2 from 3 ( 4 select ename 5 ,sal 6 ,dense_rank() over (order by sal desc) ranking 7 from emp 8 ) 9 where ranking = 4  Replace 4 with any value of N 10 / ENAME SAL RANKING    BLAKE 2850 4 CLARK 2850 4
The 4th position has a tie between BLAKE and CLARK.
Scenario 2: RANK () for Nth highest row, gaps in case of ties
The analytic function rank() will rank the rows with gaps in ranking sequence if there are ties.
The ranks are calculated as:
SQL> select ename 2 ,sal 3 ,rank() over (order by sal desc) ranking 4 from emp; ENAME SAL RANKING    KING 5000 1 FORD 3000 2 SCOTT 3000 2 JONES 2975 4 CLARK 2850 5 BLAKE 2850 5 ALLEN 1600 7 TURNER 1500 8
Wrap a filter around and pick out the Nth highest salary, say the 4th highest salary.
SQL> select * 2 from 3 ( 4 select ename 5 ,sal 6 ,rank() over (order by sal desc) ranking 7 from emp 8 ) 9 where ranking = 4  Replace 4 with any value of N 10 / ENAME SAL RANKING    JONES 2975 4
A different answer from the previous query, as there is no rank 3 because of the tied 2nd place.
Closing Notes
The requirement to “find Nth highest row” is incomplete, until the following questions are also answered:
 Can the result match more than one value? If not, on what basis should the one record be chosen if there is a tie?
 How should the subsequent records be ranked in case of ties – contiguously or with gaps?
Depending on the answer for (2), DENSE_RANK (for contiguous) or RANK (for gaps) can be used. Depending on the answer for (1), extra filter criteria can be applied to the SQL.
There are other approaches for calculating the Nth highest row, too. The next is a nonanalytic approach, which works the same way as the RANK query (gaps for ties).
SQL> select ename 2 , sal 3 from emp a 4 where 3 = ( select count(*)  Replace 3 with any value of (N  1) 5 from emp b 6 where b.sal > a.sal) 7 / ENAME SAL   JONES 2975
However, tests have shown the analytics approach to be more efficient than the nonanalytics one for Nth highest or TopN type of queries.
{ 40 comments… read them below or add one }
well…thanks for helping me
very good answer
nice xplanation,,,,,very simple
The Explanations are incredible..Keep up the good work.Thanks a ton
Nice explanation….
This explanation is very easy to understand…thanx
cool examples good work
Very useful answer, I was searching for easier and understandable explaination. It works in Oracle.
Thank you very much!!!!
Thanks Dear,
Now i got the right answer after years.Thank you very much God Bless You.
Thanx a lot…….
CLEAR !! and Very useful.
Very Helpful… Thank you
Very good answers……
thanks..nice
Can anyone explain the execution process of this query.. i m unable to understand this query execution please any one help me..
select ename, sal from emp a
where 3 = ( select count(*)from emp b where b.sal > a.sal)
thanks..nice example
Terrific explanation. Thanks.
Thanks a lot…
Awesome!
Thanks For the Info, This is awesome
I have a doubt, if we use single select sql as :
select ename ,sal ,rank() over (order by sal desc) ranking from emp where ranking = 4
this should be better, or is there any issue with this query
hello jitendrea
this query will not run its will show u a warning message about ranking is invalid identifire
so you have to use this inside the braces
Thanks Alot…..
thanks and It is very help full.
Wonderful answer. Thanks a lot
I had another doubt, both the results using DENSE_RANK() and RANK() are very different.
My question is which was the right result?
awsum..that was clear explaination. .
thanks bro/madam,
it is very useful to me.
Thanku so much……..the way of demonstrating tha concept was so good and clear…….It helped me a lot……Thanks a lot again 🙂
Hi Thank you so much!!
can anyone explain that how to write a oracle query to display the employee earning the highest salary ?
Hi,
Thanks! Very simple 🙂
Hi,
To find the nth highest salary where clause is used and with where clause the the term ranking is used which is alias. But in oracle database alias can not be used with where clause…then how it is correct ans…it will give error…ORA:00904: invalid identifier..
SQL> select *
2 from
3 (
4 select ename
5 ,sal
6 ,rank() over (order by sal desc) ranking
7 from emp
8 )
9 where ranking = 4 — Replace 4 with any value of N
10 /
@Tushar Patel: Did you run the SQL to verify that claim?
The alias ‘ranking’ has been used in an inline view, not in the main SQL’s WHERE clause.
thank you… after a long time i got the nice simple answer
excellent explanation
Thanks
thanks you for sharing this. This is one of best question for interview point of view.
thanks…. Very good explanation
SELECT *
FROM emp Emp1
WHERE (3) = (
SELECT COUNT(DISTINCT(Emp2.sal ))
FROM emp Emp2
WHERE Emp2.sal > Emp1.sal )
–the above query will work same as dense_rank()