mysql 子查询高级经典例题

SELECT d.*
FROM departments d
WHERE d.department_id=(
    SELECT department_id
    FROM employees
    GROUP BY department_id
    HAVING AVG(salary) =
    (
        SELECT MIN(a)
        FROM(
            SELECT AVG(salary) a,department_id
            FROM employees
            GROUP BY department_id
                 )b
    )
)

或

法2：通过排序然后LIMIT直接找到工资最低的部门标号，然后匹配部门表

SELECT d.*
FROM departments d
WHERE d.department_id=(
    SELECT department_id
    FROM employees
    GROUP BY department_id
    ORDER BY AVG(salary) ASC
    LIMIT 1
);

查询平均工资最低的部门信息和该部门的平均工资
法：将department表和带有最低平均工资的表连接起来，再查询

SELECT d.*,a
FROM departments d
INNER JOIN(
    SELECT AVG(salary) a,department_id
    FROM employees
    GROUP BY department_id
    ORDER BY AVG(salary) ASC
    LIMIT 1
) b
ON d.department_id=b.department_id;

查询平均工资最高的job信息

SELECT *
FROM jobs
WHERE jobs.`job_id`=(
    SELECT job_id
    FROM employees e
    GROUP BY e.job_id
    ORDER BY AVG(salary) DESC
    LIMIT 1
);

查询平均工资高于公司平均工资的部门有些
法：查找平均工资高于公司平均工资的表，然后和部门表连接

    SELECT department_name
FROM departments d
INNER JOIN(
    SELECT AVG(salary),department_id
    FROM employees
    GROUP BY department_id
    HAVING AVG(salary)>(
        SELECT AVG(salary)
        FROM employees
    )
) a
WHERE d.department_id=a.department_id;

查询出公司中所有manager的详细信息在employees表中

SELECT *
FROM employees
WHERE employee_id IN(
    SELECT manager_id
    FROM employees
);

查询各个部门中最高工资中的那个部门的最低工资是多少

SELECT MIN(e.salary)
FROM employees e
WHERE e.department_id=(
    SELECT department_id
    FROM employees
    GROUP BY department_id
    ORDER BY MAX(salary) DESC
    LIMIT 1
)

查询平均工资最高的部门的manager的详细信息：last_name,department_id,email,salary

SELECT e.last_name,e.department_id,e.email,e.salary
FROM employees e
INNER JOIN departments d
ON d.manager_id=e.employee_id
WHERE d.department_id=(
    SELECT department_id
    FROM employees
    GROUP BY department_id
    ORDER BY AVG(salary) DESC
    LIMIT 1
)

mysql 子查询高级经典例题

原文：https://blog.51cto.com/14437184/2438938