sql min function and other column - mysql

Question

i want to know if it is possible to add another column to a select statement that contain an aggregate function like min, max ...
example :
SELECT user_id, MAX(salary) FROM users;
is this statement correct in the sql standard(in mysql its work );
its work in mysql, but i think i read somewhere that if i put an aggregate function in the select clause, i can't put anything but an aggregate function or if there is a group by, the grouped column can be in the select clause (in mysql)
EDIT :
User(user_id, name, last_name, salary)
i want to select the user_id, name, (maximum salary column) from the User table; is it possible to do it without sub query?
User Table
User_id, Name, Salary
| 1 | user1 | last1 | 500 | |
|---|-------|-------|------|---|
| 2 | user2 | last2 | 1000 | |
| 3 | user3 | last3 | 750 | |
| | | | | |
the output must be the user_id, username, lastname, and salary of the user who have the max salary, so here the ouput must be :
2 user2 last2 1000

Answer 1

To start with: No,
SELECT user_id, MAX(salary) FROM users;
is not standard-compliant. You are using an aggregate function (MAX) without a GROUP BY clause. By doing so you tell the DBMS to aggregate all records to one single result row. Now what do you tell the DBMS to show in this result row? The maximum salary found in the table (MAX(salary)) and the user_id. However, there is no the user_id; there are possibly many different user_id in the table. This violates the SQL standard. MySQL takes the liberty to interpret the non-aggregated user_id as any user_id (arbitrarily picked).
So even though the query runs, it's result is usually not the desired one.
This query:
SELECT user_id, name, MAX(salary) FROM users GROUP BY user_id;
on the other hand is standard-compliant. Let's see again what this query does: This time there is a GROUP BY clause telling the DBMS you want one result row per user_id. For each user_id you want to show: the user_id, the name, and the maximum salary. All these are valid expressions; the user_id is the user_id itself, the name is the one user name associated with the user_id, and the maximum salary is the user's maximum salary. The unaggregated column name is allowed, because it is functionally dependent on the grouped-by user_id. Many DBMS don't support this, though, because it can get extremely complicated to determine whether an expression is functionally dependent on the group or not.
As to how to show the user record with the maximum salary, you need a limiting clause. MySQL provides LIMIT for this, which can get you the first n rows. It doesn't deal with ties however.
SELECT * FROM users ORDER BY salary DESC LIMIT 1;
is
SELECT * FROM users ORDER BY salary FETCH FIRST ROW ONLY;
in standard SQL.
In order to deal with ties, however, as in
SELECT * FROM users ORDER BY salary FETCH FIRST ROW WITH TIES;
you need a subquery in MySQL, because LIMIT doesn't support this:
SELECT * FROM users WHERE salary = (SELECT MAX(salary) FROM users);

Answer 2

Told you there are different solutions depending on what you want....
no group by, no subquery, Easy cake
select *
from users
ORDER BY salary DESC
LIMIT 1

Answer 3

Let's look at an example:
mysql> select * from users;
+---------+----------+
| user_id | salary |
+---------+----------+
| 1 | 42000.00 |
| 2 | 39000.00 |
| 3 | 50000.00 |
+---------+----------+
mysql> SELECT user_id, MAX(salary) FROM users;
+---------+-------------+
| user_id | MAX(salary) |
+---------+-------------+
| 1 | 50000.00 |
+---------+-------------+
What's up with that? User 1 is not the user that has a salary of 50000.00.
mysql> SELECT user_id, MAX(salary), MIN(SALARY) FROM users;
+---------+-------------+-------------+
| user_id | MAX(salary) | MIN(SALARY) |
+---------+-------------+-------------+
| 1 | 50000.00 | 39000.00 |
+---------+-------------+-------------+
User 1 is also not the one with 39000.00. This is getting fishy, right?
When you use aggregate functions, they only apply to the column you use the function in. The user_id column doesn't magically know which row that max value came from, and show the corresponding user_id.
In that example, I query both the MAX and MIN salary. But these belong to different users! Which user_id should be shown, even if the user_id could automatically be from the row where the aggregate value comes from?
And what if two users have the same salary, which are tied for the max salary? Which user_id should be displayed?
And what if you use an aggregate function that doesn't return a value that exists on any single row?
mysql> SELECT user_id, AVG(salary) FROM users;
+---------+--------------+
| user_id | AVG(salary) |
+---------+--------------+
| 1 | 43666.666667 |
+---------+--------------+
Here's the explanation: an aggregate function causes the result to be reduced to one single row, after reading the whole group of rows. A column that is not inside an aggregate function (like user_id here) takes its value from some arbitrary row in the group of rows. Arbitrary does not mean random—in practice, it tends to be the first MySQL row reads in the group. But there's no guarantee that'll always be the case.
How useful is this? Not very. In other databases, it's not a valid query, and it will literally generate an error.
In fact, MySQL 5.7 changed the behavior, by enforcing a rule that disallows ambiguous queries. If you try to run the query above on MySQL 5.7, it'll generate an error:
ERROR 1140 (42000): In aggregated query without GROUP BY, expression #1 of SELECT list contains nonaggregated column 'test.users.user_id'; this is incompatible with sql_mode=only_full_group_by
There's an option to make it act like earlier versions of MySQL. For more information on this, read: https://dev.mysql.com/doc/refman/5.7/en/group-by-handling.html
As a matter of trivia, SQLite is another database that allows this kind of arbitrary result. Only in SQLite, the value of user_id would come from the last row read in the group. Go figure.

Answer 4

Try to use this:
SELECT id,
salary
FROM (SELECT id,
salary,
MAX(salary) over ([partition by] [order by] dept) mx_sal
FROM your_tbl)
WHERE salary = mx_sal;