[英]Grouping by aggregate function max in MySQL
我有桌教练,如下所示:
+-------+------------+------------+----------+
| ID | name | dept_name | salary |
+-------+------------+------------+----------+
| 10101 | Srinivasan | Comp. Sci. | 65000.00 |
| 12121 | Wu | Finance | 90000.00 |
| 15151 | Mozart | Music | 40000.00 |
| 22222 | Einstein | Physics | 95000.00 |
| 32343 | El Said | History | 60000.00 |
| 33456 | Gold | Physics | 87000.00 |
| 45565 | Katz | Comp. Sci. | 75000.00 |
| 58583 | Califieri | History | 62000.00 |
| 76543 | Singh | Finance | 80000.00 |
| 76766 | Crick | Biology | 72000.00 |
| 83821 | Brandt | Comp. Sci. | 92000.00 |
| 98345 | Kim | Elec. Eng. | 80000.00 |
+-------+------------+------------+----------+
查询
select dept_name, max(salary) from instructor group by dept_name;
会给我教员每个部门的最高薪水。
但是,我想获取结果中的所有列,即每个部门中薪资最高的讲师的ID和名称 。 但是我不确定该怎么做。
您可以将当前查询用作过滤原始表的子查询:
SELECT i1.*
FROM instructor i1
INNER JOIN
(
SELECT dept_name, MAX(salary) AS salary
FROM instructor
GROUP BY dept_name
) i2
ON i1.dept_name = i2.dept_name AND i1.salary = i2.salary;
请使用以下查询:
SELECT ID, name, dept_name, salary
FROM( select ID, name, dept_name, salary, ROW_NUMBER() OVER(PARTITION BY dept_name ORDER BY salary DESC) as ranks
from instructor)T
WHERE ranks = 1
您还可以使用相关子查询
select t.* from instructor t
where salary in
(
select max(salary) from instructor t1
where t1.dept_name=t.dept_name
group by t1.dept_name
)
SELECT ins.id , ins.name
FROM instructor ins
inner join (
SELECT ins2.id as id , MAX(ins2.salary) as salary
FROM instructor ins2
group by ins2.id
) as insG
ON ins.id = insG.id AND ins.salary = insG.salary
您可以使用相关子查询:
select i.*
from instructor i
where salary = (select max(i1.salary)
from instructor i1
where i1.dept_name = i.dept_name
);
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