LeetCode——Department Highest Salary(花式使用IN以及GROUP BY)
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2023-11-09 21:19:52
以前使用 ,都是局限于单个数值使用,从未尝试过多个数据使用 . 此题涉及两个表,肯定需要使用 操作. 此外,需要选取每个 的最大数值,那么肯定涉及 以及 操作. 综合以上因素,答案如下所示: ......
the employee table holds all employees. every employee has an id, a salary, and there is also a column for the department id. +----+-------+--------+--------------+ | id | name | salary | departmentid | +----+-------+--------+--------------+ | 1 | joe | 70000 | 1 | | 2 | jim | 90000 | 1 | | 3 | henry | 80000 | 2 | | 4 | sam | 60000 | 2 | | 5 | max | 90000 | 1 | +----+-------+--------+--------------+ the department table holds all departments of the company. +----+----------+ | id | name | +----+----------+ | 1 | it | | 2 | sales | +----+----------+ write a sql query to find employees who have the highest salary in each of the departments. for the above tables, your sql query should return the following rows (order of rows does not matter). +------------+----------+--------+ | department | employee | salary | +------------+----------+--------+ | it | max | 90000 | | it | jim | 90000 | | sales | henry | 80000 | +------------+----------+--------+
以前使用in
,都是局限于单个数值使用,从未尝试过多个数据使用in
.
此题涉及两个表,肯定需要使用join
操作.
此外,需要选取每个department
的最大数值,那么肯定涉及max
以及group by
操作.
综合以上因素,答案如下所示:
# write your mysql query statement below select employee.name as employee, employee.salary, department.name as department from employee, department where employee.departmentid = department.id and (employee.departmentid, employee.salary) in (select employee.departmentid, max(employee.salary) from employee group by employee.departmentid);