select a.master_id,sum(case when b.status=6 then 1 else 0 end) as total_count from a inner join b on a.order_id=b.id group by a.master_id.

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');

第一种用法:

SELECT name,

CASE WHEN birthday < '1981' THEN 'old'

WHEN birthday > '1988' THEN 'yong'

ELSE 'ok' END YORN

FROM lee

第二种用法：

SELECT NAME, CASE name

WHEN 'sam' THEN 'yong'

WHEN 'lee' THEN 'handsome'

ELSE 'good' END as oldname

FROM lee

第三种：当然了，case when 语句还可以复合

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，要不然可能结果和预期的结果不同，

当然也可以用year函数来实现

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);

第一题：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

select payment_date, amount,

case

when amount >= 0 AND amount < 40 then 'low'

when amount >=40 AND amount < 80 then 'moderate'

when amount >=80 then 'high'

else 'null' END

FROM penalties

第二题:统计出属于low的罚款编号

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'