表的建立 mysql
CREATE TABLE `lee` (
`id` int(10) NOT NULL AUTO_INCREMENT,
`name` char(20) DEFAULT NULL,
`birthday` datetime DEFAULT NULL,
PRIMARY KEY (`id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8 sql
数据插入： 函数
insert into lee(name,birthday) values ('sam','1990-01-01'); spa
insert into lee(name,birthday) values ('lee','1980-01-01'); table
insert into lee(name,birthday) values ('john','1985-01-01'); date
使用case when语句 select
1。 sql语句
select name,
case
when birthday<'1981' then 'old'
when birthday>'1988' then 'yong'
else 'ok' END YORN
from lee; 方法
2。 im
select NAME,
case name
when 'sam' then 'yong'
when 'lee' then 'handsome'
else 'good' end
from lee;
固然了case when语句还能够复合
3。
select name,birthday,
case
when birthday>'1983' then 'yong'
when name='lee' then 'handsome'
else 'just so so ' end
from lee;
在这里用sql语句进行日期比较的话，须要对年加引号。要否则可能结果可能和预期的结果会不一样。个人mysql版本5.1
固然也能够用year函数来实现，以第一个sql为例
select NAME,
CASE
when year(birthday)>1988 then 'yong'
when year(birthday)<1980 then 'old'
else 'ok' END
from lee;
create table penalties
(
paymentno INTEGER not NULL,
payment_date DATE not null,
amount DECIMAL(7,2) not null,
primary key(paymentno)
)
insert into penalties values(1,'2008-01-01',3.45);
insert into penalties values(2,'2009-01-01',50.45);
insert into penalties values(3,'2008-07-01',80.45);
1.#对罚款登记分为三类，第一类low，包括大于0小于等于40的罚款，第二类moderate大于40
#到80之间的罚款，第三类high包含全部大于80的罚款。
2.#统计出属于low的罚款编号。
第一道题的解法与上面的相同
select paymentno,amount,
case
when amount>0 and amount<=40 then 'low'
when amount>40 and amount<=80 then 'moderate'
when amount>80 then 'high'
else 'incorrect' end lvl
from `penalties`
2.#统计出属于low的罚款编号。重点看这里的解决方法
方法1.
select paymentno,amount
from `penalties`
where case
when amount>0 and amount<=40 then 'low'
when amount>40 and amount<=80 then 'moderate'
when amount>80 then 'high'
else 'incorrect' end ='low';
方法2 select * from (select paymentno,amount, case when amount>0 and amount<=40 then 'low' when amount>40 and amount<=80 then 'moderate' when amount>80 then 'high' else 'incorrect' end lvl from `penalties`) as p where p.lvl='low';