开发者

How can I obtain the model's name or the content type of a Django object?

Let's say I am in the save code. How can I obtain the model's name or the 开发者_如何学编程content type of the object, and use it?

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:

import django.db.models
from django.contrib.contenttypes.models import ContentType

content_type = ContentType.objects.get(model=model_name)
model = content_type.model_class()


You can get the model name from the object like this:

self.__class__.__name__

If you prefer the content type, you should be able to get that like this:

from django.contrib.contenttypes.models import ContentType
ContentType.objects.get_for_model(self)


The method get_for_model does 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_name in:

content_type = ContentType.objects.get(model=model_name)

Use Foo._meta.model_name, or if you have a Foo object, then obj._meta.model_name is what you're looking for. Then, you can do things like

Bar.objects.filter(content_type__model=Foo._meta.model_name)

This is an efficient way to filter the Bar table to return you objects which link to the Foo content type via a field named content_type.


Using gravelpot's answer, to give a direct answer to the OP's question:

We can get the object's class through instance.__class__ and then pass it to the get_for_model function:

from django.contrib.contenttypes.models import ContentType
content_type = ContentType.objects.get_for_model(instance.__class__)
0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜