mysql 如何将sum的和重复显示到每一个数据当中

[qq_34626915]

例如：
date name neo amt
2013/04/18 AAA B1 1
2013/04/18 bbb b1 2
2013/04/17 ccc b1 3
2013/04/18 ddd b2 4
2013/04/18 eee b2 5

将相同的neo的amt值SUM以后，要得到
date name neo amt
2013/04/18 AAA B1 1
2013/04/18 bbb b1 5
2013/04/17 ccc b1 5
2013/04/18 ddd b2 9
2013/04/18 eee b2 9

gruop by neo 以后 每个neo就只有一条数据了

[qq_34626915]

没有人么QAQ

[qq_34626915]

Z

[敬以直内]

用子查询join

[Bai_jf]

select *, sum() over (partition by neo)
from

[Bai_jf]

sum(amt) over (partition by neo)

[gikod]

select date, name, neo, (select sum(amt) from t b where b.neo = a.neo) amt from t a

[ACMAIN_CHM]

引用 4 楼 gikod 的回复:

select date, name, neo, (select sum(amt) from t b where b.neo = a.neo) amt from t a

正解。

[ACMAIN_CHM]

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