Python Django - モデルフィールドのForeignKeyクラスの選択

Question

モデルの作成時とインスタンス時に、フィールドモデル/クラスの外部キーを選択できるようにしたいと思います。したがって、このような何か：

class Posts(models.Model): 

    OwnerType = (
     (Accounts, 'User'), 
     (Groups, 'Group'), 
    ) 

    PostTypes = (
     ('Text', 'Text'), 
     ('Photo', 'Photo'), 
     ('Audio', 'Audio'), 
     ('Video', 'Video'), 
     ('Link', 'Link'), 
    ) 

    owner = models.ForeignKey(OwnerType, default = 0, related_name = "post_owner") 
    ownerid = models.IntegerField(blank = False, default = 0) 
    owner_type = models.CharField(blank = False, default = '', max_length = 50) 

    title = models.CharField(max_length = 500, default = '') 
    contents = models.CharField(max_length = 500, default = '') 
    attachment = models.CharField(max_length = 500, default = '') 
    attachment_type = models.CharField(max_length = 500, default = '') 
    link = models.CharField(max_length = 500, default = '') 

    post_type = models.CharField(max_length = 20, choices = PostTypes, default = '') 
    status = models.CharField(max_length = 20, default = 'public') # either public, private, or deleted 

    date_created = models.DateField(auto_now_add=True) 
    last_active = models.DateField(auto_now=True) 

    @property 
    def serialize(self): 
     return { 
      'p_id': self.id, 

      'ownerid': self.ownerid, 
      'owner': self.owner.serialize, 
      'owner_type': self.owner_type, 
      'title': self.title, 
      'contents': self.contents, 
      'attachment': self.attachment, 
      'attachment_type': self.attachment_type, 
      'link': self.link, 
      'post_type': self.post_type, 
      'status': self.status, 
      'date_created': self.date_created, 
      'last_active': self.last_active 
     } 

    class Meta: 
     db_table = "posts" 

# --- 

しかし、私はこのエラーを取得：

Traceback (most recent call last): 
    File "manage.py", line 9, in <module> 
    execute_from_command_line(sys.argv) 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 353, in execute_from_command_line 
    utility.execute() 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 327, in execute 
    django.setup() 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/__init__.py", line 18, in setup 
    apps.populate(settings.INSTALLED_APPS) 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/apps/registry.py", line 108, in populate 
    app_config.import_models(all_models) 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/apps/config.py", line 202, in import_models 
    self.models_module = import_module(models_module_name) 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py", line 37, in import_module 
    __import__(name) 
    File "/Users/RyanWaite/Desktop/epsity/webapp/models.py", line 320, in <module> 
    class Posts(models.Model): 
    File "/Users/RyanWaite/Desktop/epsity/webapp/models.py", line 335, in Posts 
    owner = models.ForeignKey(OwnerType, default = 0, related_name = "post_owner") 
    File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/models/fields/related.py", line 754, in __init__ 
    RECURSIVE_RELATIONSHIP_CONSTANT, 
AssertionError: ForeignKey(((<class 'webapp.models.Accounts'>, u'User'), (<class 'webapp.models.Groups'>, u'Group'))) is invalid. First parameter to ForeignKey must be either a model, a model name, or the string u'self' 

はあっても可能ですか？モデルの新しいインスタンスを作成するときに、外部キーのモデルを選択できますか？投稿を作成できるモデルには、ユーザーとグループの2種類があります。ユーザは投稿を行うことができ、グループは投稿を行うことができる。ドキュメントから

Answer 1

doc

A normal ForeignKey can only “point to” one other model, which means that if the TaggedItem model used a ForeignKey it would have to choose one and only one model to store tags for

あなたは関係がどのモデルであることを可能にするGenericForeignKeyをので、ここで

Answer 2

このまわりで私のカスタムの方法である使用することができます私はタイプとTYPE_IDを追加

ここで、typeはモデルクラスの文字列です。 私は、与えられたparamsに基づいてモデルインスタンスを返す関数を作成しました。例えば、 ：私は、シリアル化プロパティのメソッドを呼び出すとき

def returnModelSerialized(type, id): 
    if id == None: 
     return 'error --- Missing id' 

    if type == '' or Type == None: 
     return 'error --- invalid type' 

    if type == 'Account': 
     obj = Accounts.objects.get(id = id) 
     return obj.serialize 

    if type == 'Group': 
     obj = Groups.objects.get(id = id) 
     return obj.serialize 

ので、それは次のようになります：

class Likes(models.Model): 

OwnerType = (
    ('Account', 'Account'), 
    ('Group', 'Group'), 
) 

ContentType = (
    ('Post', 'Post'), 
    ('Comment', 'Comment'), 
    ('Reply', 'Reply'), 
    ('Group', 'Group'), 
    ('Event', 'Event'), 
) 

ownerid = models.IntegerField(blank = False, default = 0) 
owner_type = models.CharField(choices = OwnerType, blank = False, default = '', max_length = 50) 

item_type = models.CharField(choices = ContentType, blank = False, default = '', max_length = 50) 
item_id = models.IntegerField(blank = False, default = 0) 

date_created = models.DateField(auto_now_add=True) 
last_active = models.DateField(auto_now=True) 

@property 
def serialize(self): 
    return { 
     'like_id': self.id, 

     'ownerid': self.ownerid, 
     'owner': returnModelSerialized(self.owner_type , self.ownerid), 
     'owner_type': self.owner_type, 
     'item_type': self.item_type.serialize, 
     'item_id': self.item_id, 
     'date_created': self.date_created, 
     'last_active': self.last_active 
    } 

class Meta: 
    db_table = "likes"