# 第09章_子查询
# 1. 由一个具体的需求,引入子查询
# 需求:谁的工资比Abel的高?
#方式1:
SELECT last_name, salary
FROM employees
WHERE last_name = 'Abel';
SELECT last_name, salary
FROM employees
WHERE salary > 11000;
#方式2:自连接
SELECT e2.last_name, e2.salary
FROM employees e1,
employees e2
WHERE e2.salary > e1.salary
AND e1.last_name = 'Abel';
SELECT e2.last_name, e2.salary
FROM employees e1
JOIN employees e2
ON e2.salary > e1.salary
WHERE e1.last_name = 'Abel';
#方式3: 子查询
SELECT last_name, salary
FROM employees
WHERE salary > (
SELECT salary
FROM employees
WHERE last_name = 'Abel'
);
# 2. 称谓的规范: 外查询(或主查询)、内查询(或子查询)
/*
子查询(内查询)在主查询之前一次执行完成。
子查询的结果被主查询(外查询)使用 。
注意事项
子查询要包含在括号内
将子查询放在比较条件的右侧
单行操作符对应单行子查询,多行操作符对应多行子查询
*/
#不推荐
SELECT last_name, salary
FROM employees
WHERE (
SELECT salary
FROM employees
WHERE last_name = 'Abel'
) < salary;
/*
# 3. 子查询的分类
角度1:从内查询返回的结果的条目数
单行子查询 vs 多行子查询
角度2:内查询是否被执行多次
相关子查询 vs 不相关子查询
比如:相关子查询的需求:查询工资大于本部门平均工资的员工信息。
不相关子查询的需求:查询工资大于本公司平均工资的员工信息。
*/
#子查询的编写技巧(或步骤):① 从里往外写 ② 从外往里写
# 4. 单行子查询
# 4.1 单行操作符:= != > >= < <=
# 题目:查询工资大于149号员工工资的员工的信息
SELECT employee_id, last_name, salary
FROM employees
WHERE salary > (
SELECT salary
FROM employees
WHERE employee_id = 149
);
# 题目:返回job_id与141号员工相同,salary比143号员工多的员工姓名,job_id和工资
SELECT last_name, job_id, salary
FROM employees
WHERE job_id = (
SELECT job_id
FROM employees
WHERE employee_id = 141
)
AND salary > (
SELECT salary
FROM employees
WHERE employee_id = 143
);
# 题目:返回公司工资最少的员工的last_name,job_id和salary
SELECT last_name, job_id, salary
FROM employees
WHERE salary = (
SELECT MIN(salary)
FROM employees
);
# 题目:查询与141号员工的manager_id和department_id相同的其他员工
# 的employee_id,manager_id,department_id。
# 方式1:
SELECT employee_id, manager_id, department_id
FROM employees
WHERE manager_id = (
SELECT manager_id
FROM employees
WHERE employee_id = 141
)
AND department_id = (
SELECT department_id
FROM employees
WHERE employee_id = 141
)
AND employee_id <> 141;
# 方式2:了解
SELECT employee_id, manager_id, department_id
FROM employees
WHERE (manager_id, department_id) = (
SELECT manager_id, department_id
FROM employees
WHERE employee_id = 141
)
AND employee_id <> 141;
# 题目:查询最低工资大于110号部门最低工资的部门id和其最低工资
SELECT department_id, MIN(salary)
FROM employees
WHERE department_id IS NOT NULL
GROUP BY department_id
HAVING MIN(salary) > (
SELECT MIN(salary)
FROM employees
WHERE department_id = 110
);
# 题目:显式员工的employee_id,last_name和location。
# 其中,若员工department_id与location_id为1800的department_id相同,
# 则location为’Canada’,其余则为’USA’。
SELECT employee_id,
last_name,
CASE department_id
WHEN (SELECT department_id FROM departments WHERE location_id = 1800) THEN 'Canada'
ELSE 'USA' END "location"
FROM employees;
# 4.2 子查询中的空值问题
SELECT last_name, job_id
FROM employees
WHERE job_id = (SELECT job_id
FROM employees
WHERE last_name = 'Haas'
);
# 4.3 非法使用子查询
#错误:Subquery returns more than 1 row
SELECT employee_id, last_name
FROM employees
WHERE salary =(
SELECT MIN(salary)
FROM employees
GROUP BY department_id
);
# 5. 多行子查询
# 5.1 多行子查询的操作符: IN ANY ALL SOME(同ANY)
# 5.2 举例
# IN:
SELECT employee_id, last_name
FROM employees
WHERE salary IN (
SELECT MIN(salary)
FROM employees
GROUP BY department_id
);
# ANY / ALL:
# 题目:返回其它job_id中比job_id为‘IT_PROG’部门任一工资低的员工的员工号、
# 姓名、job_id 以及salary
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE job_id <> 'IT_PROG'
AND salary < ANY (
SELECT salary
FROM employees
WHERE job_id = 'IT_PROG'
);
# 题目:返回其它job_id中比job_id为‘IT_PROG’部门所有工资低的员工的员工号、
# 姓名、job_id 以及salary
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE job_id <> 'IT_PROG'
AND salary < ALL (
SELECT salary
FROM employees
WHERE job_id = 'IT_PROG'
);
# 题目:查询平均工资最低的部门id
#MySQL中聚合函数是不能嵌套使用的。
#方式1:
SELECT department_id
FROM employees
GROUP BY department_id
HAVING AVG(salary) = (
SELECT MIN(avg_sal)
FROM(
SELECT AVG(salary) avg_sal
FROM employees
GROUP BY department_id
) t_dept_avg_sal
);
#方式2:
SELECT department_id
FROM employees
GROUP BY department_id
HAVING AVG(salary) <= ALL (
SELECT AVG(salary) avg_sal
FROM employees
GROUP BY department_id
);
# 5.3 空值问题
SELECT last_name
FROM employees
WHERE employee_id NOT IN (
SELECT manager_id
FROM employees
WHERE manager_id IS NOT NULL
);
# 6. 相关子查询
# 回顾:查询员工中工资大于本公司平均工资的员工的last_name,salary和其department_id
# 6.1
SELECT last_name,salary,department_id
FROM employees
WHERE salary > (
SELECT AVG(salary)
FROM employees
);
# 题目:查询员工中工资大于本部门平均工资的员工的last_name,salary和其department_id
# 方式1:使用相关子查询
SELECT last_name,salary,department_id
FROM employees e1
WHERE salary > (
SELECT AVG(salary)
FROM employees e2
WHERE department_id = e1.department_id
);
# 方式2:
SELECT e.last_name,e.salary,e.department_id
FROM employees e ,(
SELECT department_id,AVG(salary) avg_sal
FROM employees
GROUP BY department_id
) t_dept_avg_sal
WHERE e.department_id = t_dept_avg_sal.department_id
AND e.salary > t_dept_avg_sal.avg_sal;
# 题目:查询员工的id,salary,按照department_name 排序
SELECT employee_id,salary
FROM employees e
ORDER BY (
SELECT department_name
FROM departments d
WHERE e.department_id = d.department_id
) ASC;
# 结论:在SELECT中,除了GROUP BY 和 LIMIT之外,其他位置都可以声明子查询!
/*
SELECT ....,....,....(存在聚合函数)
FROM ....(LEFT / RIGHT)JOIN....ON 多表的连接条件
(LEFT / RIGHT)JOIN .... ON ....
WHERE 不包含聚合函数的过滤条件
GROUP BY ....,....
HAVING 包含聚合函数的过滤条件
ORDER BY ....,....(ASC / DESC)
LIMIT ....,....
*/
# 题目:若employees表中employee_id与job_history表中employee_id相同的数目不小于2,
# 输出这些相同id的员工的employee_id,last_name和其job_id
SELECT *
FROM job_history;
SELECT employee_id,last_name,job_id
FROM employees e
WHERE 2 <= (
SELECT COUNT(*)
FROM job_history j
WHERE e.employee_id = j.employee_id;
# 6.2 EXISTS 与 NOT EXISTS 关键字
# 题目:查询公司管理者的employee_id,last_name,job_id,department_id信息
#方式1:自连接
SELECT DISTINCT mgr.employee_id,mgr.last_name,mgr.job_id,mgr.department_id
FROM employees emp JOIN employees mgr
ON emp.manager_id = mgr.employee_id;
#方式2:子查询
SELECT employee_id,last_name,job_id,department_id
FROM employees
WHERE employee_id IN (
SELECT DISTINCT manager_id
FROM employees
);
#方式3:使用EXISTS
SELECT employee_id,last_name,job_id,department_id
FROM employees e1
WHERE EXISTS(
SELECT *
FROM employees e2
WHERE e1.employee_id = e2.manager_id
);
# 题目:查询departments表中,不存在于employees表中的部门的department_id和department_name
# 方式1:
SELECT d.department_id,d.department_name
FROM employees e RIGHT JOIN departments d
ON e.department_id = d.department_id
WHERE e.department_id IS NULL;
# 方式2:
SELECT department_id,department_name
FROM departments d
WHERE NOT EXISTS(
SELECT *
FROM employees e
WHERE d.department_id = e.department_id
);
SELECT COUNT(*)
FROM departments;
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