Python 如何获取模型的名称或 Django 对象的内容类型？
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How can I obtain the model's name or the content type of a Django object?
提问by Seitaridis
Let's say I am in the save code. How can I obtain the model's name or the content type of the object, and use it?
假设我在保存代码中。如何获取模型的名称或对象的内容类型并使用它？
from django.db import models
class Foo(models.Model):
...
def save(self):
I am here....I want to obtain the model_name or the content type of the object
This code works, but I have to know the model_name:
此代码有效，但我必须知道模型名称：
import django.db.models
from django.contrib.contenttypes.models import ContentType
content_type = ContentType.objects.get(model=model_name)
model = content_type.model_class()
采纳答案by gravelpot
You can get the model name from the object like this:
您可以像这样从对象中获取模型名称：
self.__class__.__name__
If you prefer the content type, you should be able to get that like this:
如果你更喜欢内容类型，你应该能够得到这样的：
ContentType.objects.get_for_model(self)
回答by AlanSE
The method get_for_modeldoes some fancy stuff, but there are some cases when it's better to not use that fancy stuff. In particular, say you wanted to filter a model that linked to ContentType, maybe via a generic foreign key?? The question here was what to use for model_namein:
该方法get_for_model做了一些花哨的东西，但在某些情况下最好不要使用那些花哨的东西。特别是，假设您想过滤链接到 ContentType 的模型，可能通过通用外键？这里的问题是用于什么model_name：
content_type = ContentType.objects.get(model=model_name)
content_type = ContentType.objects.get(model=model_name)
Use Foo._meta.model_name, or if you have a Fooobject, then obj._meta.model_nameis what you're looking for. Then, you can do things like
使用Foo._meta.model_name，或者如果您有一个Foo对象，那么obj._meta.model_name这就是您要查找的内容。然后，您可以执行以下操作
Bar.objects.filter(content_type__model=Foo._meta.model_name)
This is an efficient way to filter the Bartable to return you objects which link to the Foocontent type via a field named content_type.
这是过滤Bar表以返回Foo通过名为 的字段链接到内容类型的对象的有效方法content_type。
