Django(1.10)覆盖AdminSite

Question

我试图用我自己的自定义类覆盖AdminSite类.我按照django的文档中的教程:https://docs.djangoproject.com/en/1.10/ref/contrib/admin/#customizing-adminsite但它没有用.具体来说,我想用我自己的类覆盖原始的AdminSite,而不只是在我的项目中添加另一个管理站点.

我创建了从类继承的自定义类MyAdminSite

from django.contrib.admin import AdminSite


class MyAdminSite(AdminSite):
    pass

然后在我的app urls.py中添加:

from django.conf.urls import url, include
import django.contrib.admin as admin
from .admin_site import MyAdminSite

admin.site = MyAdminSite()
admin.autodiscover()


urlpatterns = [
    url(r'^', admin.site.urls),
]

它似乎工作,但管理员模型注册到AdminAdite的MyAdminSite.

我尝试了三种注册模型到我的自定义网站:

@admin.register(Model)
class ModelAdmin(model.AdminModel):
...

这种方式模型已注册到原始AdminSite.

第二种方式:

@admin.site.register(Model):
class ModelAdmin(model.AdminModel):
...

这不起作用并导致异常.ModelAdmin类未传递给register方法.

最后一种方式

class ModelAdmin(model.AdminModel):
...
admin.site.register(Model, ModelAdmin)

这是有效的,但在管理员网站上我只能看到我的模型不是来自Django管理员(用户和组)的模型.

如何永久覆盖admin.site并将所有模型注册到MyAdminSite？

Answer 1

来自myapp/admin.py:

from django.contrib.auth.models import Group, User
from django.contrib.auth.admin import GroupAdmin, UserAdmin
from django.contrib.admin import AdminSite
from django.contrib import admin

from .models import MyModel #This is my app's model

# Custom admin site
class MyAdminSite(AdminSite):
    site_header = 'My Project Title'
    site_title  = 'My Project Title Administration'
    index_title = 'My Project Title Administration'
    # You can add on more attributes if you need 
    # Check out https://docs.djangoproject.com/en/1.11/ref/contrib/admin/#adminsite-objects

# Create admin_site object from MyAdminSite
admin_site = MyAdminSite(name='my_project_admin')

# Register the models
class MyModelAdmin(admin.ModelAdmin):
    list_display = ('id', 'description')

admin_site.register(MyModel, MyModelAdmin)

# Create and register all of your models
# ....

# This is the default Django Contrib Admin user / group object
# Add this if you need to edit the users / groups in your custom admin
admin_site.register(Group, GroupAdmin)
admin_site.register(User, UserAdmin)

来自myproject/urls.py

from django.conf.urls import url
from django.contrib import admin
from myapp.admin import admin_site ##! Important..Import your object (admin_site) instead of your class (MyAdminSite)

urlpatterns = [
    url(r'^admin/', admin_site.urls), #Now all /admin/ will go to our custom admin
]

Answer 2

我没有找到解决问题的方法，但已采取了解决方法。

首先，我们需要在应用程序中创建模块（例如admin.py），然后扩展AdminSite类：

from django.contrib.admin import AdminSite

class MyAdminSite(AdminSite):
    ...

然后，在模块底部，我们需要创建MyAdminSite的实例并从Django注册内置模型：

site = MyAdminSite()
site.register(Group, GroupAdmin)
site.register(User, UserAdmin)

必要进口：

from django.contrib.auth.models import Group, User
from django.contrib.auth.admin import GroupAdmin, UserAdmin

在我们的网站网址模块中，我们需要覆盖原始网站对象：

from .admin import site

admin.site = site
admin.autodiscover()
...
url(r'^admin/', admin.site.urls)
...

我们需要做的最后更改是注册我们的模型。我们需要记住的一件事是，我们不能像这样使用register作为装饰器：

@admin.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
    ...

要么：

@admin.site.register(MyModel)
class MyModelAdmin(admin.ModelAdmin):
    ...

我们需要定义我们的ModelAdmin类，然后在MyAdminSite对象上调用register：

class MyModelAdmin(admin.ModelAdmin):
    ...
admin.site.register(MyModel, MyModelAdmin)

这是唯一对我有用的解决方案。