我有这个模型:
class Movie(models.Model):
# I use taggit for tag management
tags = taggit.managers.TaggableManager()
class Person(models.Model):
# manytomany with a intermediary model
movies = models.ManyToManyField(Movie, through='Activity')
class Activity(models.Model):
movie = models.ForeignKey(Movie)
person = models.ForeignKey(Person)
name = models.CharField(max_length=30, default='actor')我想匹配一部演员和另一部电影演员相同的电影。Not one actor in common, but all the actors in common。
所以我不想这样:
# actors is a shortcut property
one_actor_in_common = Movie.object.filter(activities__name='actor',
team_members__in=self.movie.actors)我想要一些能让“黑客帝国I”和“黑客帝国II”匹配的东西,因为他们共享“基努里维斯”和“劳伦斯菲什伯恩”,但不能匹配“速度”,因为他们共享“基努里维斯”而不是“劳伦斯菲什伯恩”。
发布于 2012-01-05 20:46:03
多对多的管理器不能同时匹配多个关系。在数据库级别,这一切都归结为选择和分组。
因此,自然地说,数据库能够回答的唯一问题是:列出涉及这些人的表演活动,按电影对它们进行分组,并仅显示那些具有与人相同数量的表演活动的电影。
翻译成ORM语言,它看起来像这样:
actors = Person.objects.filter(name__in=('Keanu Reaves', 'Laurence Fishburne'))
qs = Movie.objects.filter(activity__name='actor',
activity__person__in=actors)
qs = qs.annotate(common_actors=Count('activity'))
all_actors_in_common = qs.filter(common_actors=actors.count())这样产生的查询实际上还不错:
SELECT "cmdb_movie"."id", "cmdb_movie"."title", COUNT("cmdb_activity"."id") AS "common_actors"
FROM "cmdb_movie"
LEFT OUTER JOIN "cmdb_activity" ON ("cmdb_movie"."id" = "cmdb_activity"."movie_id")
WHERE ("cmdb_activity"."person_id" IN (SELECT U0."id" FROM "cmdb_person" U0 WHERE U0."name" IN ('Keanu Reaves', 'Laurence Fishburne'))
AND "cmdb_activity"."name" = 'actor' )
GROUP BY "cmdb_movie"."id", "cmdb_movie"."title", "cmdb_movie"."id", "cmdb_movie"."title"
HAVING COUNT("cmdb_activity"."id") = 2我还有一个小应用程序,我用它来测试它,但我不知道是否有人需要它,也不知道在哪里托管它。
https://stackoverflow.com/questions/7470071
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