SQL实战（三）

标签：not   多次   employee   sql   order   图片   row   table   情况   

一、

 查找所有员工自入职以来的薪水涨幅情况，给出员工编号emp_noy以及其对应的薪水涨幅growth，并按照growth进行升序
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_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`));

employees 中tem_no和salaries中tem_no插入，然后找最大最小，求值

1、

select sCurrent.emp_no,(sCurrent.salary-sStart.salary) as growth
from (select e.emp_no,s.salary from employees as e,salaries as s where e.emp_no=s.emp_no and s.to_date="9999-01-01") as sCurrent  --找到当前薪水
join (select e.emp_no,s.salary from employees as e,salaries as s where e.emp_no=s.emp_no and s.from_date=e.hire_date) as sStart  --找到初始入职薪水
on sCurrent.emp_no=sStart.emp_no   --合并表，相减
order by growth 

生成新表，两个新表连接后再减，而不能在一个表里多次跨行减。　　

2、

更好理解

select a.emp_no,(b.salary-c.salary) as growth
from employees as a  --往该表添加
join salaries as b on a.emp_no=b.emp_no and b.to_date="9999-01-01"   --加入当前薪水
join salaries as c on a.emp_no=c.emp_no and c.from_date=a.hire_date  --加入最初薪水
order by growth

二、

SQL实战（三）

标签：not   多次   employee   sql   order   图片   row   table   情况   

原文地址：https://www.cnblogs.com/ruo-li-suo-yi/p/8977088.html