MYSQL query to find the all employees with nth highest salary - mysql

Question

The two tables are salary_employee and employee
employee_salary
salary_id emp_id salary
Employee
emp_id | first_name | last_name | gender | email | mobile | dept_id | is_active
Query to get the all employees who have nth highest salary where n =1,2,3,... any integer
SELECT a.salary, b.first_name
FROM employee_salary a
JOIN employee b
ON a.emp_id = b.emp_id
WHERE a.salary = (
SELECT salary
FROM employee_salary
GROUP BY salary
DESC
LIMIT 1 OFFSET N-1
)
My Questions:
1) Is there any better and optimized way we can query this,
2) Is using LIMIT an good option
3) We have more options to calculate the nth highest salary, which is the best and what to follow and when?
One option using :
SELECT *
FROM employee_salary t1
WHERE ( N ) = ( SELECT COUNT( t2.salary )
FROM employee_salary t2
WHERE t2.salary >= t1.salary
)
Using Rank Method
SELECT salary
FROM
(
SELECT #rn := #rn + 1 rn,
a.salary
FROM tableName a, (SELECT #rn := 0) b
GROUP BY salary DESC
) sub
WHERE sub.rn = N

Answer 1

You have asked what seems like a reasonable question. There are different ways of doing things in SQL and sometimes some methods are better than others. The ranking problem is just one of many, many examples. The "answer" to your question is that, in general, order by is going to perform better than group by in MySQL. Although even that depends on the particular data and what you consider to be "better".
The specific issues with the question are that you have three different queries that return three different things.
The first returns all employees with a "dense rank" that is the same. That terminology is use purposely because it corresponds to the ANSI dense_rank() function which MySQL does not support. So, if your salaries are 100, 100, and 10, it will return two rows with a ranking of 1 and one with a ranking of 2.
The second returns different results if there are ties. If the salaries are 100, 100, 10, this version will return no rows with a ranking of 1, two rows with a ranking of 2, and one row with a ranking of 3.
The third returns an entirely different result set, which is just the salaries and the ranking of the salaries.
My comment was directed at trying the queries on your data. In fact, you should decide what you actually want, both from a functional and a performance perspective.

Answer 2

LIMIT requires the SQL to skim through all records between 0 and N and therefore requires increasing time the further back in your ranking you want to look. However, IMO that problem cannot be solved better.
As Gordon Linoff suggested: Run your option against your data set, using the commonly used ranks (which ranks are queried often, which are not? The result might be fast on rank 1 but terrible on rank 100).
Execute and analyze the Query Execution Plan and create indexes accordingly (for example on the salary column) and retest your queries.
Other options:
Option 4:
You could build a ranking table whichs serves as cache. The execution plan of your Limit-Query shows (see sqlfiddle here), that mysql already does create a temporary table to solve the query.
Pros: Easy and fast
Cons: Forces you to regenerate the ranking table each time the data changes
Option 5:
You could reconsider how you define "ranks".
If we have the following salaries:
100'000
100'000
80'000
Is the employee Nr 3 considered to be of rank 3 or 2?
Are 1 and 2 on the same rank (rank 1), but 3 is on rank 3?
If you define rank = order, you can greatly simplify the query to
SELECT a.salary, b.first_name
FROM employee_salary a, employee b
WHERE a.emp_id = b.emp_id
order by salary desc
LIMIT 1 OFFSET 4
demo: http://sqlfiddle.com/#!2/e7321d/1/0

Answer 3

try this,
SELECT * FROM one as A WHERE ( n ) = ( SELECT COUNT(DISTINCT(b.salary)) FROM one as B WHERE
B.salary >= A.salary )

Answer 4

Suppose emp_salary table have the below records:
And you want to select all employees with nth (N=1,2,3 etc.) highest/lowest (only change >(for highest), < (for lowest) operator according to your needs) salary, use the below sql:
SELECT DISTINCT(a.salary),
a.id,
a.name
FROM emp_salary a
WHERE N = (SELECT COUNT( DISTINCT(b.salary)) FROM emp_salary b
WHERE b.salary >= a.salary
);
For example, if you want to select all employees with 2nd highest salary, use below sql:
SELECT DISTINCT(a.salary),
a.id,
a.name
FROM emp_salary a
WHERE 2 = (SELECT COUNT( DISTINCT(b.salary)) FROM emp_salary b
WHERE b.salary >= a.salary
);
But if you want to display only second highest salary(only single record), use the below sql:
SELECT DISTINCT(a.salary),
a.id,
a.name
FROM emp_salary a
WHERE 2 = (SELECT COUNT( DISTINCT(b.salary)) FROM emp_salary b
WHERE b.salary >= a.salary
) limit 1;