MySQL 8.0

By using db<>fiddle, you agree to license everything you submit by Creative Commons CC0.

select version();


version()


8.0.27


create table mytable(id int, salary int, managerid int);
insert into mytable (id, salary, managerid) values
(1, 1000, null), (2, 1000, null), (3, 900, null), (4, 900, 2),
(5, 800, 1), (6, 800, 2), (7, 700, 1), (8, 600, 2), (9, 500, 3);
select *
from
(
  select t.*, dense_rank() over (order by salary desc) as rnk
  from mytable t
) employee
left join mytable manager on manager.id = employee.managerid
where employee.rnk = 3;


id
salary
managerid
rnk
id
salary
managerid


5
800
1
3
1
1000
null

6
800
2
3
2
1000
null


select *
from mytable employee
left join mytable manager on manager.id = employee.managerid
where employee.salary =
(
  select distinct salary
  from mytable
  order by salary desc
  limit 2, 1
);


id
salary
managerid
id
salary
managerid


5
800
1
1
1000
null

6
800
2
2
1000
null