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 Models Group By
• Django equivalent for count and group by
• How to query as GROUP BY in django?
• How to use the ORM for the equivalent of a SQL count, group and join query?

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this widely upvoted and discussed meta answer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

我们可以在Django ORM上执行GROUP BY ... COUNT或GROUP BY ... SUM SQL等效查询，使用annotate()，values()，django.db.models的Count和Sum方法，并且可选地使用order_by()方法：

• GROUP BY ... COUNT：

  result = Books.objects.values('author')
                        .order_by('author')
                        .annotate(count=Count('author'))

  现在结果包含一个包含两个键的字典：author和count：

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

• GROUP BY ... 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
      ...     |      ...