题目描述

获取有奖金的员工相关信息。

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 `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 emp_bonus(
emp_no int not null,
received datetime not null,
btype smallint not null);

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`));

给出emp_no、first_name、last_name、奖金类型btype、对应的当前薪水情况salary以及奖金金额bonus。 bonus类型btype为1其奖金为薪水salary的10%，btype为2其奖金为薪水的20%，其他类型均为薪水的30%。 当前薪水表示to_date=’9999-01-01’

输入描述

无

输出描述

emp_no	first_name	last_name	btype	salary	bonus
10001	Georgi	Facello	1	88958	8895.8
10002	Bezalel	Simmel	2	72527	14505.4
10003	Parto	Bamford	3	43311	12993.3
10004	Chirstian	Koblick	1	74057	7405.7

题解

SELECT e.emp_no, e.first_name, e.last_name, eb.btype, s.salary, 
(s.salary * eb.btype / 10.0) as bonus
from employees as e inner join emp_bonus as eb on e.emp_no=eb.emp_no
inner join salaries as s on s.emp_no=e.emp_no
and s.to_date='9999-01-01';

备注：

1. 使用SQLite 中 CASE 表达式

   SELECT e.emp_no, e.first_name, e.last_name, b.btype, s.salary, 
   (CASE b.btype 
    WHEN 1 THEN s.salary * 0.1
    WHEN 2 THEN s.salary * 0.2
    ELSE s.salary * 0.3 END) AS bonus
   FROM employees AS e INNER JOIN emp_bonus AS b ON e.emp_no = b.emp_no
   INNER JOIN salaries AS s ON e.emp_no = s.emp_no AND s.to_date = '9999-01-01'

参考文章：

参考 [ciphersaw] 的解答