汇总各个部门当前员工的title类型的分配数目

题目描述

汇总各个部门当前员工的title类型的分配数目，结果给出部门编号dept_no、dept_name、其当前员工所有的title以及该类型title对应的数目count
CREATE TABLE `departments` (
`dept_no` char(4) NOT NULL,
`dept_name` varchar(40) NOT NULL,
PRIMARY KEY (`dept_no`));
CREATE TABLE `dept_emp` (
`emp_no` int(11) NOT NULL,
`dept_no` char(4) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`dept_no`));
CREATE TABLE IF NOT EXISTS `titles` (
`emp_no` int(11) NOT NULL,
`title` varchar(50) NOT NULL,
`from_date` date NOT NULL,
`to_date` date DEFAULT NULL);

输入描述:

无

输出描述:

dept_no	dept_name	title	count
d001	Marketing	Senior Engineer	1
d001	Marketing	Staff	1
d002	Finance	Senior Engineer	1
d003	Human Resources	Senior Staff	1
d004	Production	Senior Engineer	2
d005	Development	Senior Staff	1
d006	Quality Management	Engineer	2
d006	Quality Management	Senior Engineer	1

代码：

select ddd.dept_no,dd.dept_name,dt.title,count(dt.title) as count
from titles as dt inner join dept_emp as ddd
on ddd.emp_no = dt.emp_no and ddd.to_date='9999-01-01' and dt.to_date = '9999-01-01'
inner join departments as dd
on dd.dept_no = ddd.dept_no
group by dd.dept_no,dt.title;

如果只按dept_no分组，那么每个dept_no只会出现一条记录，则不能体现同一dept_no中的不同title。GROUP BY de.dept_no, t.title 的作用是同时将de.dept_no和t.title分组，即要两者都相同才能判别为相同的分组，只要一个不同就算不同的分组。