题目描述
统计各个部门对应员工涨幅的次数总和,给出部门编码dept_no、部门名称dept_name以及次数sum
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 `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
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 `salaries` (
`emp_no` int(11) NOT NULL,
`salary` int(11) NOT NULL,
`from_date` date NOT NULL,
`to_date` date NOT NULL,
PRIMARY KEY (`emp_no`,`from_date`));
输入描述:
无
输出描述:
dept_no | dept_name | sum |
---|---|---|
d001 | Marketing | 24 |
d002 | Finance | 14 |
d003 | Human Resources | 13 |
d004 | Production | 24 |
d005 | Development | 25 |
d006 | Quality Management | 25 |
select de.dept_no,dp.dept_name,count(s.salary) as sum
from (dept_emp as de inner join salaries as s on de.emp_no = s.emp_no)
inner join departments as dp
on de.dept_no = dp.dept_no
group by de.dept_no;
本题关键是要将 每个部门分组,并分别统计工资记录总数,思路如下:
1、用INNER JOIN连接dept_emp表和salaries表,并以dept_emp.no分组,统计每个部门所有员工工资的记录总数
2、再将上表用INNER JOIN连接departments表,限制条件为两表的dept_no相等,找到dept_no与dept_name的对应关系,最后依次输出dept_no、dept_name、sum