django - count(*) 在 Django Orm 中

Question

我找到了这个，但我需要类似的东西

select count(*) as my_count from my_table;

在 Django 中可能吗？

谢谢。

Answer 1

例如，如果您有“Book”对象，只需转换 .count() 方法：

Book.objects.count()

Answer 2

假设您要进行聚合。Django ORM 可以做到。这是一个文档。

简而言之，您可以汇总一些计数，然后在查询中使用它。文档中的示例：

class Author(models.Model):
   name = models.CharField(max_length=100)
   age = models.IntegerField()

class Publisher(models.Model):
   name = models.CharField(max_length=300)
   num_awards = models.IntegerField()

class Book(models.Model):
   name = models.CharField(max_length=300)
   pages = models.IntegerField()
   price = models.DecimalField(max_digits=10, decimal_places=2)
   rating = models.FloatField()
   authors = models.ManyToManyField(Author)
   publisher = models.ForeignKey(Publisher)
   pubdate = models.DateField()

class Store(models.Model):
   name = models.CharField(max_length=300)
   books = models.ManyToManyField(Book)
   registered_users = models.PositiveIntegerField()

现在，当你这样做时：

>>> from django.db.models import Count
>>> pubs = Publisher.objects.annotate(num_books=Count('book'))
>>> pubs

SQL查询将如下：

SELECT "main_publisher"."id", "main_publisher"."name", "main_publisher"."num_awards", COUNT("main_book"."id") AS "num_books" FROM "main_publisher"LEFT OUTER JOIN "main_book" ON ("main_publisher"."id" = "main_book"."publisher_id") GROUP BY "main_publisher"."id", "main_publisher"."name", "main_publisher"."num_awards" LIMIT 21; args=()

您可以在过滤中使用聚合数据：

>>> pubs = Publisher.objects.annotate(num_books=Count('book')).filter(num_books=2)
>>> pubs

SELECT "main_publisher"."id", "main_publisher"."name", "main_publisher"."num_awards", COUNT("main_book"."id") AS "num_books" FROM "main_publisher"LEFT OUTER JOIN "main_book" ON ("main_publisher"."id" = "main_book"."publisher_id") GROUP BY "main_publisher"."id", "main_publisher"."name", "main_publisher"."num_awards" HAVING COUNT("main_book"."id") = 2  LIMIT 21; args=(2,)