如何在Django ORM中执行GROUP BY ... COUNT或SUM？

Question

序幕:

这是SO中经常出现的问题:

• Django模型组
• Django相当于count和group by
• 如何在django中查询GROUP BY？
• 如何使用ORM相当于SQL计数,组和连接查询？

我已经在SO文档中编写了一个示例,但由于文档将于2017年8月8日关闭,我将遵循这个广泛上升和讨论的元答案的建议,并将我的示例转换为自我回答的帖子.

当然,我也很乐意看到任何不同的方法!

---

题:

假设模型:

class Books(models.Model):
    title  = models.CharField()
    author = models.CharField()
    price = models.FloatField()

如何使用Django ORM在该模型上执行以下查询:

• GROUP BY ... COUNT:

  SELECT author, COUNT(author) AS count
  FROM myapp_books GROUP BY author
  

• GROUP BY ... SUM:

  SELECT author,  SUM (price) AS total_price
  FROM myapp_books GROUP BY author
  

Answer 1

我们可以进行一个GROUP BY ... COUNT或者一个GROUP BY ... SUM在Django的ORM SQL的等效查询,使用的annotate(),values()的django.db.models的Count和Sum恭敬的方法和可选的order_by()方法:

• GROUP BY ... COUNT:

  from django.db.models import Count
  
  result = Books.objects.values('author')
                        .order_by('author')
                        .annotate(count=Count('author'))
  

  现在结果包含一个带有两个键的字典:author和count:

    author    | count
  ------------|-------
   OneAuthor  |   5
  OtherAuthor |   2
     ...      |  ...
  

• GROUP BY ... SUM:

  from django.db.models import Sum
  
   result = Books.objects.values('author')
                         .order_by('author')
                         .annotate(total_price=Sum('price'))
  

  现在结果包含一个包含两列的字典:author和total_price:

    author    | total_price
  ------------|-------------
   OneAuthor  |    100.35
  OtherAuthor |     50.00
      ...     |      ...