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MySQL---Subqueries

2022-07-31 17:16:00 【Narwhals need water】

子查询

引入

谁的工资比Abel的高？

方式1：
SELECT 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` #多表的连接条件
AND 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'
		);

子查询的基本使用

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

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.

单行子查询

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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
		);

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
		);

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
		);

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;

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
		     );

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;

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');

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);

多行子查询

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IN:
SELECT employee_id, last_name
FROM   employees
WHERE  salary IN
                (SELECT   MIN(salary)
                 FROM     employees
                 GROUP BY department_id); 

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'
		);

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'
		);

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
			) 

空值问题
SELECT last_name
FROM employees
WHERE employee_id NOT IN (
			SELECT manager_id
			FROM employees
			);

相关子查询

回顾：查询员工中工资大于公司平均工资的员工的last_name,salary和其department_id

SELECT last_name,salary,department_id
FROM employees
WHERE salary > (
		SELECT AVG(salary)
		FROM employees
		);

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：在FROM中声明子查询
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


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 ...,....


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`
		)

EXISTS 与 NOT EXISTS关键字

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`
	     );


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;

相关更新

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相关删除

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思考题

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https://yzsam.com/2022/212/202207311656432731.html

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