mysql
语句case when

select USER_ID ,USER_NAME ,CASE WHEN  atten.DESTINATION_ID is null THEN FALSE ELSE TRUE END  as attentioned    from T_SD_USER as user LEFT OUTER JOIN T_SD_ATTENTION as atten on `user`.USER_ID = atten.ORIGIN_ID and atten.DESTINATION_ID = 3001 where `user`.USER_ID = 3000;

表的创建

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

数据插入：

insert into lee(name,birthday) values (‘sam‘,‘1990-01-01‘);

insert into lee(name,birthday) values (‘lee‘,‘1980-01-01‘);
insert into lee(name,birthday) values (‘john‘,‘1985-01-01‘);

使用case when语句

1。

select name,

case

when birthday<‘1981‘ then ‘old‘

when birthday>‘1988‘ then ‘yong‘

else ‘ok‘ END YORN

from lee;

2。

select NAME,

case name

when ‘sam‘ then ‘yong‘

when ‘lee‘ then ‘handsome‘

else ‘good‘ end

from lee;

当然了case when语句还可以复合

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，需要对年加引号。要不然可能结果可能和预期的结果会不同。我的mysql版本5.1

当然也可以用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包含所有大于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‘;

来自 http://www.cnblogs.com/john2000/archive/2010/09/21/1832729.html