题目：数据库中存在如下数据，求用户最终余额。

Sum(Decode(...))能解决以上同类所有的问题，详细的我就不写了，你自己写的基于会深刻些！

如有更好的方法请各位留个言！

同求MySQL的最优实现方法，因为接下来将要使用MySQL数据库了！

3楼Myue昨天 13:48

[code=sql]nnWITH t (用户 ,类型, 金额 ) ASn(select 'A','存入',100 UNION ALL nselect 'A','存入', 200 UNION ALL nselect 'A','取出', 100 UNION ALL nselect 'A', '取出', 200 UNION ALL nselect 'A','存入' ,300 UNION ALL nselect 'A','取出', 300 )n,t1 asn(SELECT * ,ROW_NUMBER() OVER (ORDER BY 用户 )AS row1 FROM t )n,t2 AS n(SELECT *,ROW_NUMBER() OVER (ORDER BY row1 )+1 AS row2 FROM t1)n,t3 AS (SELECT * ,金额1=CASE 类型 WHEN '存入' THEN 金额 ELSE -金额 end FROM t2)nSELECT *, (SELECT SUM(金额1) FROM t3 n WHERE n.row1< w.row2 ) AS 余额 FROM t3 AS w[/code]

Re: bkq4215115856小时前

回复Myuen亲，如果有1000条数据，难不成要写1000个select？？？

2楼F172515968昨天 10:34

select 用户 as 用户, sum(金额) as 余额 from (select 用户 as 用户,金额 as 金额 from tb where 类型号‘存入’ union all select 用户 as 用户,-金额as 金额 from tb where 类型号‘取出’ ) t group by 用户

1楼adyfang昨天 09:52

SUM(IF(类型='取出', -金额, 金额))