正如jpic和sgallen在评论中所建议的那样，可以.filter()为每个类别添加一个选项。每filter增加一个，就会添加更多的联接，这对于少量的类别来说应该不是问题。
有聚合 方法。对于大量类别，此查询将更短，甚至更快。
您还可以选择使用自定义查询。
一些例子
测试设置：
class Photo(models.Model):
tags = models.ManyToManyField(‘Tag’)
class Tag(models.Model):
name = models.CharField(max_length=50)
def __unipre__(self): return self.name
In [2]: t1 = Tag.objects.create(name=’holiday’)
In [3]: t2 = Tag.objects.create(name=’summer’)
In [4]: p = Photo.objects.create()
In [5]: p.tags.add(t1)
In [6]: p.tags.add(t2)
In [7]: p.tags.all()
Out[7]: [
使用链接过滤器方法：
In [8]: Photo.objects.filter(tags=t1).filter(tags=t2)
Out[8]: [
结果查询：
In [17]: print Photo.objects.filter(tags=t1).filter(tags=t2).query
SELECt “test_photo”.”id”
FROM “test_photo”
INNER JOIN “test_photo_tags” ON (“test_photo”.”id” = “test_photo_tags”.”photo_id”)
INNER JOIN “test_photo_tags” T4 ON (“test_photo”.”id” = T4.”photo_id”)
WHERe (“test_photo_tags”.”tag_id” = 3 AND T4.”tag_id” = 4 )
请注意，每个都为查询filter添加了更多内容JOINS。
使用注释 方法：
In [29]: from django.db.models import Count
In [30]: Photo.objects.filter(tags__in=[t1, t2]).annotate(num_tags=Count(‘tags’)).filter(num_tags=2)
Out[30]: [
结果查询：
In [32]: print Photo.objects.filter(tags__in=[t1, t2]).annotate(num_tags=Count(‘tags’)).filter(num_tags=2).query
SELECt “test_photo”.”id”, COUNT(“test_photo_tags”.”tag_id”) AS “num_tags”
FROM “test_photo”
LEFT OUTER JOIN “test_photo_tags” ON (“test_photo”.”id” = “test_photo_tags”.”photo_id”)
WHERe (“test_photo_tags”.”tag_id” IN (3, 4))
GROUP BY “test_photo”.”id”, “test_photo”.”id”
HAVINg COUNT(“test_photo_tags”.”tag_id”) = 2
ANDed Q对象不起作用：
In [9]: from django.db.models import Q
In [10]: Photo.objects.filter(Q(tags__name=’holiday’) & Q(tags__name=’summer’))
Out[10]: []
In [11]: from operator import and_
In [12]: Photo.objects.filter(reduce(and_, [Q(tags__name=’holiday’), Q(tags__name=’summer’)]))
Out[12]: []
结果查询：
In [25]: print Photo.objects.filter(Q(tags__name=’holiday’) & Q(tags__name=’summer’)).query
SELECt “test_photo”.”id”
FROM “test_photo”
INNER JOIN “test_photo_tags” ON (“test_photo”.”id” = “test_photo_tags”.”photo_id”)
INNER JOIN “test_tag” ON (“test_photo_tags”.”tag_id” = “test_tag”.”id”)
WHERe (“test_tag”.”name” = holiday AND “test_tag”.”name” = summer )
