八：子查询（where、having）
#进阶七：子查询
/*
说明：当一个查询语句中又嵌套了另一个完整的select语句，则被嵌套的select语句称为子查询或内查询
外面的select语句称为主查询或外查询。
分类：
按子查询出现的位置进行分类：
1、select后面
要求：子查询的结果为单行单列（标量子查询）
2、from后面
要求：子查询的结果可以为多行多列
3、where或having后面 ★
要求：子查询的结果必须为单列
单行子查询
多行子查询
4、exists后面
要求：子查询结果必须为单列（相关子查询）
特点：
1、子查询放在条件中，要求必须放在条件的右侧
2、子查询一般放在小括号中
3、子查询的执行优先于主查询
4、单行子查询对应了 单行操作符：> < >= <= = <>
多行子查询对应了 多行操作符：any/some all in
#一、放在where或having后面
#一）单行子查询
#案例1：谁的工资比 Abel 高?
#①查询Abel的工资
SELECT salary
FROM employees
WHERE last_name = 'Abel'
#②查询salary>①的员工信息
SELECT last_name,salary
FROM employees
WHERE salary>(
SELECT salary
FROM employees
WHERE last_name = 'Abel'
);
#案例2：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
#①查询141号员工的job_id
SELECT job_id
FROM employees
WHERE employee_id = 141
#②查询143号员工的salary
SELECT salary
FROM employees
WHERE employee_id = 143
③查询job_id=① and salary>②的信息
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
);
#案例3：返回公司工资最少的员工的last_name,job_id和salary
#①查询最低工资
SELECT MIN(salary)
FROM employees
#②查询salary=①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
);
#案例4：查询最低工资大于50号部门最低工资的部门id和其最低工资
#①查询50号部门的最低工资
SELECT MIN(salary)
FROM employees
WHERE department_id = 50
#②查询各部门的最低工资，筛选看哪个部门的最低工资>①
SELECT MIN(salary),department_id
FROM employees
GROUP BY department_id
HAVING MIN(salary)>(
SELECT MIN(salary)
FROM employees
WHERE department_id = 50
);
练习：
#1. 查询和 Zlotkey 相同部门的员工姓名和工资
#①查询Zlotkey的部门编号
SELECT department_id
FROM employees
WHERE last_name = 'Zlotkey'
#②查询department_id = ①的员工姓名和工资
SELECT last_name,salary
FROM employees
WHERE department_id = (
SELECT department_id
FROM employees
WHERE last_name = 'Zlotkey'
);
#2. 查询工资比公司平均工资高的员工的员工号，姓名和工资。
#①查询平均工资
SELECT AVG(salary)
FROM employees
#②查询salary>①的信息
SELECT employee_id,last_name,salary
FROM employees
WHERE salary>(
SELECT AVG(salary)
FROM employees
);
#二）多行子查询
/*
in:判断某字段是否在指定列表内
x in(10,30,50)
any / some:判断某字段的值是否满足其中任意一个
x>any(10,30,50) 等价于 x>min()
x=any(10,30,50) 等价于 x in(10,30,50)
all:判断某字段的值是否满足里面所有的
x >all(10,30,50) 等价于 x >max()
*/
#案例1：返回location_id是1400或1700的部门中的所有员工姓名
#①查询location_id是1400或1700的部门
SELECT department_id
FROM departments
WHERE location_id IN(1400,1700)
#②查询department_id = ①的姓名
SELECT last_name 员工姓名
FROM employees
WHERE department_id IN(
SELECT DISTINCT department_id
FROM departments
WHERE location_id IN(1400,1700)
);
#题目：返回其它部门中比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
#①查询job_id为‘IT_PROG’部门的工资
SELECT DISTINCT salary
FROM employees
WHERE job_id = 'IT_PROG'
#②查询其他部门的工资<任意一个①的结果
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<ANY(
SELECT DISTINCT salary
FROM employees
WHERE job_id = 'IT_PROG'
);
等价于
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<(
SELECT MAX(salary)
FROM employees
WHERE job_id = 'IT_PROG'
);
#案例3：返回其它部门中比job_id为‘IT_PROG’部门所有工资都低的员工 的员工号、姓名、job_id 以及salary
#①查询job_id为‘IT_PROG’部门的工资
SELECT DISTINCT salary
FROM employees
WHERE job_id = 'IT_PROG'
#②查询其他部门的工资<所有①的结果
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<ALL(
SELECT DISTINCT salary
FROM employees
WHERE job_id = 'IT_PROG'
);
等价于
SELECT employee_id,last_name,job_id,salary
FROM employees
WHERE salary<(
SELECT MIN(salary)
FROM employees
WHERE job_id = 'IT_PROG'
);
#二、放在select后面
#案例；查询部门编号是50的员工个数
#方式一;
SELECT COUNT(*)
FROM `employees`
WHERE `department_id`=50;
#方式二：
SELECT
(
SELECT COUNT(*)
FROM employees
WHERE department_id = 50
) 个数;
#三、放在from后面
#案例：查询每个部门的平均工资的工资级别
#①查询每个部门的平均工资
SELECT AVG(salary),department_id
FROM employees
GROUP BY department_id
#②将①和sal_grade两表连接查询
SELECT dep_ag.department_id,dep_ag.ag,g.grade
FROM sal_grade g
JOIN (
SELECT AVG(salary) ag,department_id
FROM employees
GROUP BY department_id
) dep_ag ON dep_ag.ag BETWEEN g.min_salary AND g.max_salary;
#四、放在exists后面
#案例1 ：查询有无名字叫“Abel”的员工信息
SELECT EXISTS(
SELECT *
FROM employees
WHERE last_name = 'Abel'
) 有无Abel;
#案例2：查询没有女朋友的男神信息
USE girls;
SELECT bo.*
FROM boys bo
WHERE bo.`id` NOT IN(
SELECT boyfriend_id
FROM beauty b
)