一、分组查询

#进阶5:分组查询
/*
语法:
	select 分组函数,列(要求出现在group by的后面)
	from 表
	【where 筛选条件】
	group by 分组的列表
	【order by 子句】
注意:
	查询列表必须特殊,要求是分组函数和group by后出现的字段

特点:
	1、分组查询中的筛选条件分为两类
			数据源			位置			关键字
	分组前筛选	原始表			group by子句的前面	where
	分组后筛选	分组后的结果集		group by子句的后面	having
	
	①分组函数做条件肯定是放在having子句中
	②能用分组前筛选的,就优先考虑使用分组前筛选
	2、group by子句支持单个字段分组,多个字段分组(多个字段之间用逗号隔开没有顺序要求),表达式或函数(用得较少)
	3、也可以添加排序(排序放在整个分组查询的最后)
*/

#引入:查询每个部门的平均工资
SELECT AVG(salary) FROM employees;

#案例1:查询每个工种的最高工资
SELECT MAX(salary),job_id
FROM employees
GROUP BY job_id;

#案例2:查询每个位置上的部门个数
SELECT COUNT(*),location_id
FROM departments
GROUP BY location_id;

#添加筛选条件
#案例1:查询邮箱中包含a字符的,每个部门的平均工资
SELECT AVG(salary),department_id
FROM employees
WHERE email LIKE '%a%'
GROUP BY department_id;

#案例2:查询有奖金的每个领导手下员工的最高工资
SELECT MAX(salary),manager_id
FROM employees
WHERE commission_pct IS NOT NULL
GROUP BY manager_id;

#添加分组后的筛选条件

#案例1:查询哪个部门的员工个数>2

#①查询每个部门的员工个数
SELECT COUNT(*),department_id
FROM employees
GROUP BY department_id;

#②根据①的结果进行筛选,查询哪个部门的员工个数>2

SELECT COUNT(*),department_id
FROM employees
GROUP BY department_id  #在筛选后的列表中进行筛选,用having连接
HAVING COUNT(*)>2;


#案例2:查询每个工种有奖金的员工的最高工资>12000的工种编号和最高工资

#①查询每个工种有奖金的员工的最高工资

SELECT MAX(salary),job_id
FROM employees
WHERE commission_pct IS NOT NULL
GROUP BY job_id;

#②根据①结果继续筛选,最高工资>12000

SELECT MAX(salary),job_id
FROM employees
WHERE commission_pct IS NOT NULL
GROUP BY job_id
HAVING MAX(salary)>12000;


#案例3:查询领导编号>102的每个领导手下的最低工资>5000的领导编号是哪个,以及其最低工资

#①查询每个领导手下的员工固定最低工资

SELECT MIN(salary),manager_id
FROM employees
GROUP BY manager_id;

#②添加筛选条件:编号>102

SELECT MIN(salary),manager_id
FROM employees
WHERE manager_id>102
GROUP BY manager_id;

#③添加筛选条件:最低工资>5000

SELECT MIN(salary),manager_id
FROM employees
WHERE manager_id>102
GROUP BY manager_id
HAVING MIN(salary)>5000;

#按表达式或函数分组

#案例:按员工姓名的长度分组,查询每一组的员工个数,筛选员工个数>5的有哪些

#①查询每个长度的员工个数

SELECT COUNT(*),LENGTH(last_name) len_name
FROM employees
GROUP BY LENGTH(last_name)

#②添加筛选条件

SELECT COUNT(*),LENGTH(last_name) len_name
FROM employees
GROUP BY LENGTH(last_name)
HAVING COUNT(*)>5;

#使用别名(group by和having都支持别名,where不支持)
SELECT COUNT(*) c,LENGTH(last_name) len_name
FROM employees
GROUP BY len_name
HAVING c;


#按多个字段分组

#案例:查询每个部门每个工种的员工的平均工资

SELECT AVG(salary),department_id,job_id
FROM employees
GROUP BY department_id,job_id;
#这里group by后面两个字段颠倒顺序,不影响结果


#添加排序
#案例:查询每个部门每个工种的员工的平均工资,并且按平均工资的高低显示
SELECT AVG(salary),department_id,job_id
FROM employees
GROUP BY department_id,job_id
ORDER BY AVG(salary) DESC; #按平均工资降序

#添加条件:部门编号不为空,平均工资大于15000
SELECT AVG(salary),department_id,job_id
FROM employees
WHERE department_id IS NOT NULL
GROUP BY department_id,job_id
HAVING AVG(salary)>15000
ORDER BY AVG(salary) DESC; #按平均工资降序


二、测试题

#测试题
#1.查询各个job_id的员工工资的最大值,最小值,平均值,总和,并按job_id升序
SELECT MAX(salary),MIN(salary),AVG(salary),SUM(salary),job_id
FROM employees
GROUP BY job_id
ORDER BY job_id;

#2.查询员工最高工资和最低工资的差距(DIFFERENCE)
SELECT MAX(salary)-MIN(salary) DIFFERENCE
FROM employees;

#3.查询各个管理者手下员工的最低工资,其中最低工资不能低于600,没有管理者的员工不计算在内
SELECT MIN(salary),manager_id
FROM employees
WHERE manager_id IS NOT NULL
GROUP BY manager_id
HAVING MIN(salary)>=6000;

#4.查询所有部门的编号,员工数量和工资平均值,并按平均工资降序
SELECT department_id,COUNT(*),AVG(salary)
FROM employees
GROUP BY department_id
ORDER BY AVG(salary) DESC;

#5.选择具有各个job_id的员工人数
SELECT COUNT(*),job_id
FROM employees
GROUP BY job_id;


Logo

开放原子开发者工作坊旨在鼓励更多人参与开源活动,与志同道合的开发者们相互交流开发经验、分享开发心得、获取前沿技术趋势。工作坊有多种形式的开发者活动,如meetup、训练营等,主打技术交流,干货满满,真诚地邀请各位开发者共同参与!

更多推荐