做一个模型的所有对象提供给另一个模型

Question

如果我有以下型号：

class Player(models.Model): 
    game = models.ForeignKey('Game', null=True, blank=True) 

class Game(models.Model): 
    score = models.IntegerField(default=0) 

这样，如果我有一个类型的游戏的多个对象，p.game（其中p为类型的对象玩家）将永远是无。如果我添加n个Game类型的对象，我希望它们对所有玩家都可用，但每个玩家都应该有他自己的分数。我怎样才能做到这一点？ Django 1.4.2如果很重要。

Answer 1

我认为Django的ManyToManyField是答案：

https://docs.djangoproject.com/en/dev/topics/db/examples/many_to_many/

UPDATE：分数可以额外场M2M喜欢这里：https://docs.djangoproject.com/en/dev/topics/db/models/#extra-fields-on-many-to-many-relationships

Answer 2

您可以创建第三个模型，并保持它的简单。

class Player(models.Model): 
    name = models.CharField(default="John Rambo", max_length=256) 

class Game(models.Model): 
    name = models.CharField(default="Pack-man", max_length=256) 

class Instance(models.Model): 
    player = models.ForeignKey(Player) 
    game = models.ForeignKey(Game) 
    score = models.IntegerField(default=0) 

我明白你非常初学者所以加入一个例子：

In [1]: from xyz.models import * 

In [2]: player = Player() 
In [3]: player.save() 

In [4]: game = Game() 
In [5]: game.save() 

In [6]: instance1 = Instance.objects.create(player=player, game=game, score=0) 

In [7]: player.instance_set.all() 
Out[7]: [<Instance: Instance object>] 

In [8]: game.instance_set.all() 
Out[8]: [<Instance: Instance object>] 

In [9]: all_scores_for_john_rambo = [i.score for i in player.instance_set.filter(player__name="John Rambo")] 
Out[9]: [0]