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mysql 语句case when

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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。

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select name,
case 
when birthday<1981 then old
when birthday>1988 then yong
else ok END YORN
from lee;

 
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2。

 

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select NAME,
case name
when sam then yong
when lee then handsome
else good end
from lee;
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当然了case when语句还可以复合

3。

 

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select name,birthday,
case 
when birthday>1983 then yong
when name=lee then handsome
else just so so  end
from lee;
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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

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select NAME,
CASE
when year(birthday)>1988 then yong
when year(birthday)<1980 then old
else ok END
from lee;
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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);
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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

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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`
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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.

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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;
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方法2

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

 

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mysql 语句case when

标签:des   c   style   class   blog   code   

原文地址:http://www.cnblogs.com/jorcen/p/3761280.html

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