MySQL SUM 具有相同的 ID

抱歉，这个真正简单的问题，我刚刚学习 PHP 和 MySQL，我已经在谷歌上搜索了一个多星期，但没有找到任何答案。

我创建了一个简单的财务脚本，表格如下：

table_a
aid | value
1   | 100
2   | 50
3   | 150

table_b
bid | aid | value
1   | 1   | 10
2   | 1   | 15
3   | 2   | 5
4   | 2   | 10
5   | 3   | 25
6   | 3   | 40

我想要这样的结果

No | ID | Total | Balance
1  | 1  | 10    | 90
2  | 1  | 25    | 75
3  | 2  | 5     | 45
4  | 2  | 15    | 35
5  | 3  | 25    | 125
6  | 3  | 65    | 85

有人可以帮我解决我的问题吗？

Thanks

试试这个运行总计：http://www.sqlfiddle.com/#!2/ce765/1 http://www.sqlfiddle.com/#!2/ce765/1

select  
    bid as no, value,
    @rt := if(aid = @last_id, @rt + value, value) as total,
    @last_id := aid
from table_b b, (select @rt := 0 as x, @last_id := null) as vars
order by b.bid, b.aid;

Output:

| NO | VALUE | TOTAL | @LAST_ID := AID |
|----|-------|-------|-----------------|
|  1 |    10 |    10 |               1 |
|  2 |    15 |    25 |               1 |
|  3 |     5 |     5 |               2 |
|  4 |    10 |    15 |               2 |
|  5 |    25 |    25 |               3 |
|  6 |    40 |    65 |               3 |

然后加入表A，最终查询：

select x.no, x.aid, x.value, x.total, a.value - x.total as balance
from
(
  select    
    bid as no, aid, value,
    @rt := if(aid = @last_id, @rt + value, value) as total,
    @last_id := aid
  from table_b b, (select @rt := 0 as x, @last_id := null) as vars
  order by b.bid, b.aid
) as x
join table_a a using(aid)

Output:

| NO | AID | VALUE | TOTAL | BALANCE |
|----|-----|-------|-------|---------|
|  1 |   1 |    10 |    10 |      90 |
|  2 |   1 |    15 |    25 |      75 |
|  3 |   2 |     5 |     5 |      45 |
|  4 |   2 |    10 |    15 |      35 |
|  5 |   3 |    25 |    25 |     125 |
|  6 |   3 |    40 |    65 |      85 |

现场测试：http://www.sqlfiddle.com/#!2/ce765/1 http://www.sqlfiddle.com/#!2/ce765/1

UPDATE

不依赖于列bid排序、分组的总计不会受到影响：http://www.sqlfiddle.com/#!2/6a1e6/3 http://www.sqlfiddle.com/#!2/6a1e6/3

select x.no, x.aid, x.value, x.total, a.value - x.total as balance
from
(
  select    
    @rn := @rn + 1 as no, aid, value,
    @rt := if(aid = @last_id, @rt + value, value) as total,
    @last_id := aid
  from table_b b, (select @rt := 0 as x, @last_id := null, @rn := 0) as vars
  order by b.aid, b.bid
) as x
join table_a a using(aid)

Output:

| NO | AID | VALUE | TOTAL | BALANCE |
|----|-----|-------|-------|---------|
|  1 |   1 |    10 |    10 |      90 |
|  2 |   1 |    15 |    25 |      75 |
|  3 |   1 |     7 |    32 |      68 |
|  4 |   2 |     5 |     5 |      45 |
|  5 |   2 |    10 |    15 |      35 |
|  6 |   3 |    25 |    25 |     125 |
|  7 |   3 |    40 |    65 |      85 |

现场测试：http://www.sqlfiddle.com/#!2/6a1e6/3 http://www.sqlfiddle.com/#!2/6a1e6/3