如何将Django的GenericForeignKey限制为模型列表？ [英] How can I restrict Django's GenericForeignKey to a list of models?

问题描述

有没有办法告诉django一个内容类型GenericForeignKey的模型只能指向来自预定义列表的模型？例如，我有4个模型：A，B，C，D和一个拥有GenericForeignKey的模型X。我可以告诉X只有A& B允许用于GenericForeignKey？

解决方案

例如，
您的应用程序是app和app2，应用程序中的B模型，app2中有C，D模型。
您只想看到 app.A和app.B和app2.C

  class TaggedItem（models.Model）：
 tag = models.SlugField（）
 limit = models.Q（app_label ='app'，model ='a'）| models.Q（app_label ='app'，model ='b'）| model.Q（app_label ='app2'，model ='c'）
 content_type = models.ForeignKey（ContentType，limit_choices_to = limit）
 object_id = models.PositiveIntegerField（）
 content_object = generic .GenericForeignKey（'content_type'，'object_id'）
  

在ForeignKey上使用limit_choices_to。

检查django文档的详细信息和Q对象，app_label。
你需要写正确的app_label和model。这只是代码片段

加：我觉得你写错了app_label。这可以帮助你。

  from django.contrib.contenttypes.models在ContentType.objects中为c导入ContentType 
。 all（）：
 print（c.app_label，c.model）
  

Is there a way of telling django that a model having a contenttypes GenericForeignKey can only point to models from a predefined list? For example, I have 4 models: A, B, C, D and a model X that holds a GenericForeignKey. Can I tell X that only A & B are allowed for the GenericForeignKey?

解决方案

For example, your apps are app and app2 and there are A, B models in app and there are C, D models in app2. you want to see only app.A and app.B and app2.C

    class TaggedItem(models.Model):
        tag = models.SlugField()
        limit = models.Q(app_label = 'app', model = 'a') | models.Q(app_label = 'app', model = 'b') | models.Q(app_label = 'app2', model = 'c')
        content_type = models.ForeignKey(ContentType, limit_choices_to = limit)
        object_id = models.PositiveIntegerField()
        content_object = generic.GenericForeignKey('content_type', 'object_id')

use limit_choices_to on ForeignKey.

check django docs for details and Q objects, app_label. you need to write proper app_label and model. This is just code snippet

plus: I think you write wrong app_label. This can help you.

    from django.contrib.contenttypes.models import ContentType
    for c in ContentType.objects.all():
        print(c.app_label, c.model)

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