根据其他列以不同方式对同一列求和 - Django ORM答案

【问题标题】：Sum same column in different ways depending on other column - Django ORM根据其他列以不同方式对同一列求和 - Django ORM
【发布时间】：2016-07-04 13:09:22
【问题描述】：

我有如下表赋值结构：

| employee | product | process | qty |
| Swati    | PROD1   | issue   |  60 |
| Rohit    | PROD1   | issue   |  30 |
| Rohit    | PROD2   | issue   |  40 |
| Swati    | PROD1   | receive |  40 |
| Swati    | PROD2   | issue   |  70 |

我希望每位员工的决赛桌看起来像这样（比如employee = 'Swati'）：

| product | sum_issued | sum_received
| PROD1   |         60 |           40 |
| PROD2   |         70 |            0 |

执行此操作的 SQL 查询是：

select product
     , sum(case when process='issue' then qty else 0 end) as sum_issued
     , sum(case when process='receive' then qty else 0 end) as sum_received 
  from assignment 
 where employee = 'Swati' 
 group 
    by product;

对应于这个结果的 Django 查询应该是什么？

【问题讨论】：

标签： python mysql django

【解决方案1】：

我猜您的模型名称是“Assignment”。您可以使用以下查询

from django.db.models import Case, Value, When, Sum, IntegerField, Count

result = Assignment.objects.filter(employee="Swati").values('product').annotate(
    sum_issued=Sum(
        Case(When(process='issue', then='qty'), default=Value(0), output_field=IntegerField())),
    sum_recived=Sum(Case(When(process='receive', then='qty'), default=Value(0), output_field=IntegerField()))
    )

如果打印上面的查询print result.query，结果是，

SELECT "product", SUM(CASE WHEN "process" = issue THEN "qty" ELSE 0 END) AS "sum_issued", SUM(CASE WHEN "process" = receive THEN "qty" ELSE 0 END) AS "sum_recived" FROM "assignment" WHERE "employee" = 'Swati' GROUP BY "product"

【讨论】：

我试过写一些非常相似的东西，但查询总是给我一个由 output_field 引发的 FieldError。谢谢！