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mysql 语句case when

2024-07-06 12:15:46   228人阅读

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

 


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