Question

我的数据库中有一个类似于以下内容的表

+----+----+----+---------------------+
| id | a  | b  | date_created        |
+----+----+----+---------------------+
|  1 | 22 | 33 | 2014-07-31 14:38:17 |
|  2 | 11 |  9 | 2014-07-30 14:40:19 |
|  3 |  8 |  4 | 2014-07-29 14:40:34 |
+----+----+----+---------------------+

我正在尝试编写一个从每个sum(b)中减去a的查询。但是，b中包含的sum(b)值应该只是那些早于（或同时）a的值。换句话说，查询返回的结果应该是下面显示的结果

22 - (33 + 9 + 4)
11 - (9 + 4)
 8 - (4)

是否可以在单个查询中计算出来？

Answer 1

select id, a, a - (select sum(b) 
                     from My_TABLE T2 
                    where T2.date_created <= T1.date_created)
  from MY_TABLE T1;

Answer 2

这样的事情应该有效：

select t1.a - ifnull( sum(t2.b), 0)
from myTable t1
left outer join myTable t2 on t2.date_created <= t1.date_created 
group by t1.a

请注意，该表已连接到自身以访问两组不同的信息。

编辑：

我想你可能想按照date_created分组：

select t1.date_created, t1.a - ifnull( sum(t2.b), 0)
from myTable t1
left outer join myTable t2 on t2.date_created <= t1.date_created 
group by t1.date_created, t1.a

Answer 3

另一种选择

<强> SQL Fiddle Example

SELECT
   id,
    a - (@total := @total + b) as Total
 FROM
    (SELECT *, @total:=0
     FROM  my_table
     ORDER  BY date_created asc) AS Base

Answer 4

SELECT x.*
     , x.a - SUM(y.b) 
  FROM my_table x 
  JOIN my_table y 
    ON y.date_created <= x.date_created 
 GROUP 
    BY x.id;

查询从另一列中减去一列的总和

4 个答案: