mysql 查询结果求和

1、创建三张测试表，

create table pw_memberinfo(uid int, name varchar(20))

create table pw_members(companyid int, uid int)

create table pw_memberdata(uid int, deposit int, ddeposit int, money int)

2、三张表，分别插入测试数据，

insert into pw_members values(1, 1)

insert into pw_members values(1, 2)

insert into pw_members values(1, 3)

insert into pw_memberinfo values(1, 'name_1')

insert into pw_memberinfo values(2, 'name_2')

insert into pw_memberinfo values(3, 'name_3')

insert into pw_memberdata values(1,30,50,150)

insert into pw_memberdata values(2,77,50,12)

insert into pw_memberdata values(3,44,50,82)

3、查看pw_memberdata表中的记录，select * from pw_memberdata t，

4、编写sql语句，

select * from (SELECT i.uid, sum(deposit+ddeposit+money) as allmoney

FROM pw_memberinfo i

LEFT JOIN pw_members m ON m.uid=i.uid

LEFT JOIN pw_memberdata d ON i.uid=d.uid

group by i.uid

) t where allmoney>200

求和很简单呀，套一层sql语句就可以了，中间是你的sql语句，取出来的字段最好起个别名（这里用val），临时的那个结果也给起个别名（这里用t）：

select sum ( t.val ) from

(

select count(distinct(da1.studentno)) as val from dcs_attendance da1 group by FROM_UNIXTIME(da1.inserttime,'%Y%m%d')

) t

至于你还要做其他的，得看其他的数据关系了，现有的信息没法做。

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