Question

好吧，我有一个让我无法解决的头脑。总新手:) 如果它们出现在计算中，我需要计算库存项目数量并检测负值：

inquantity | outquantity
100        |      0
10         |      0
0          |     50
0          |     100
20         |      0
0          |     80
15         |      0
100        |      0

我需要计算Quty：

inquantity | outquantity | Quty
100        |      0      | 100
10         |      0      | 110
0          |     50      | 60
0          |     100     | -40
20         |      0      | -20
0          |     80      | -100
15         |      0      | -85
100        |      0      | 15

我该怎么做？

关于Abhik的帖子：

select 
id ,
inquantity,
outquantity,
@qty:= (@qty+inquantity)-outquantity as qty 
from quantity,(select @qty:= 0 )r 
order by id;

是否有可能在productid change上重置变量@qty？

+----+-----------+------------+-------------+------+
| id | productid | inquantity | outquantity | qty  |
+----+-----------+------------+-------------+------+
|  1 |         1 |        100 |           0 | 100  |
|  2 |         1 |         10 |           0 | 110  |
|  3 |         1 |          0 |          50 |  60  |
|  4 |         1 |          0 |         100 | -40  |
|  5 |         2 |         20 |           0 |  20  |
|  6 |         2 |          0 |          80 | -60  |
|  7 |         2 |         15 |           0 | -45  |
|  8 |         3 |        100 |           0 | 100  |
+----+-----------+------------+-------------+------+

Answer 1

如您所述，您考虑以下内容id

mysql> select * from quantity ;
+------+------------+-------------+
| id   | inquantity | outquantity |
+------+------------+-------------+
|    1 |        100 |           0 |
|    2 |         10 |           0 |
|    3 |          0 |          50 |
|    4 |          0 |         100 |
|    5 |         20 |           0 |
|    6 |          0 |          80 |
|    7 |         15 |           0 |
|    8 |        100 |           0 |
+------+------------+-------------+

我们可以获得所需的结果

select 
id ,
inquantity,
outquantity,
@qty:= (@qty+inquantity)-outquantity as qty 
from quantity,(select @qty:= 0 )r 
order by id;

输出为

+------+------------+-------------+------+
| id   | inquantity | outquantity | qty  |
+------+------------+-------------+------+
|    1 |        100 |           0 |  100 |
|    2 |         10 |           0 |  110 |
|    3 |          0 |          50 |   60 |
|    4 |          0 |         100 |  -40 |
|    5 |         20 |           0 |  -20 |
|    6 |          0 |          80 | -100 |
|    7 |         15 |           0 |  -85 |
|    8 |        100 |           0 |   15 |
+------+------------+-------------+------+

Answer 2

这个想法是获得累积总和。假设您有一个提供订购信息的列，您可以使用子查询或变量执行此操作。后者应该更有效率：

select t.*, (cumei - cumeo) as diff
from (select t.*, (@i := @i + inquantity) as cumei,
              (@o := @o + outquantity) as cumeo
      from table t
           (select @i := 0, @o := 0) vars
      order by id
     ) t
where (cumei - cumeo) < 0;

SQL数量计算

2 个答案: