Django聚合文档给出了这个例子,它计算了与每个Publisher相关的Book的数量,并返回一个带有前5个出版商的查询集,其中标注了它们的图书计数(比如用这个计数追加一个新字段):
>>> pubs = Publisher.objects.annotate(num_books=Count('book')).order_by('-num_books')[:5]
>>> pubs[0].num_books
1323但我只需要计算一种特定类型的书。就像这样。
>>> pubs = Publisher.objects.annotate(num_books=Count('book__type="nonfiction"')).order_by('-num_books')[:5]有没有过滤器可以用来实现这一点,或者我需要求助于原始SQL?
上面的文档示例类似于我的实际问题,即通过系统中的医院数量来识别和量化最大的医疗集团,当系统和医院都被建模为医疗保健Entity时:
>>> biggest_systems = Entity.objects.filter(relationships__type_id=NAME_TO_TYPE_ID.get('hospital')).annotate(num_hospitals=Count('relationships')).order_by('-num_hospitals')[:5]
>>> biggest_systems[0].num_hospitals
25relationships是一个带有直通表的实体M2M字段,type_id也是Entity中的一个字段:
class Entity(models.Model):
id = models.AutoField(verbose_name='ID',primary_key=True, db_column='ID')
type_id = models.IntegerField(verbose_name='detailed type',db_column='TypeID', blank=True, editable=True, choices=ENTITYTYPE_CHOICES, help_text="A category for this healthcare entity.")
relationships = models.ManyToManyField('self', verbose_name='affiliated with', through='EntityRelationship', symmetrical=False, related_name='related_to+')发布于 2012-08-24 13:44:18
请参阅Order of annotate() and filter() clauses
Publisher.objects.filter(book__rating__gt=3.0).annotate(num_books=Count('book'))或者在你的例子中:
Publisher.objects.filter(book__type='nonfiction').annotate(num_books=Count('book'))https://stackoverflow.com/questions/12102843
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