How can I select the record with the 2nd highest salary in database Oracle?

Question

Suppose I have a table employee with id, user_name, salary. How can I select the record with the 2nd highest salary in Oracle?

I googled it, find this solution, is t

Answer 1

I would suggest following two ways to implement this in Oracle.

1. Using Sub-query:

select distinct SALARY   
from EMPLOYEE e1  
where 1=(select count(DISTINCT e2.SALARY) from EMPLOYEE e2 where         
  e2.SALARY>e1.SALARY);

This is very simple query to get required output. However, this query is quite slow as each salary in inner query is compared with all distinct salaries.

2. Using DENSE_RANK():

select distinct SALARY   
from
  (
    select e1.*, DENSE_RANK () OVER (order by SALARY desc) as RN 
    from EMPLOYEE e
  ) E
 where E.RN=2;

This is very efficient query. It works well with DENSE_RANK() which assigns consecutive ranks unlike RANK() which assigns next rank depending on row number which is like olympic medaling.

Difference between RANK() and DENSE_RANK(): https://oracle-base.com/articles/misc/rank-dense-rank-first-last-analytic-functions