13518219792
你试试 我也不清楚了
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select sum (分数) from+表名+where +序号
in (select max(序号) from +表名+ group by+姓名)
select sum([value]-[fee]*100) from pos where user=1
一般用SQL语句求和的格式是：
select sum(column_name) from db_name where search_condition
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 allmoney200
select sum(money) as money, id
from recharge
where time between xxx and xxx
group by id
