[英]Remove default apps from Django-admin
By default, in Django-admin there is Users, Groups, and Sites apps. 默认情况下,在Django-admin中有用户,组和站点应用程序。 How can I remove Groups and Sites?
如何删除群组和网站?
I tried to remove admin.autodiscover()
from root urls. 我试图从根网址中删除
admin.autodiscover()
。 Then, when I added something like admin.site.register(User, UserAdmin)
somewhere in my app models I got an AlreadyRegistered
exception (this is fairly right - models users already registered in django.contrib.auth). 然后,当我在我的应用程序模型中的某处添加了
admin.site.register(User, UserAdmin)
类的东西时admin.site.register(User, UserAdmin)
我得到了一个AlreadyRegistered
异常(这是相当正确的 - 模型用户已在django.contrib.auth中注册)。
In an admin.py you know will definitely be loaded, try: 在admin.py中,你知道肯定会加载,尝试:
admin.site.unregister(User)
admin.site.unregister(Group)
admin.site.unregister(Site)
In addition to the above double check your ordering of "INSTALLED_APPS" in "settings.py" 除上述内容外,请仔细检查“settings.py”中“INSTALLED_APPS”的排序
INSTALLED_APPS = [
# django apps first
'django.contrib.admin',
'django.contrib.auth',
'django.contrib.contenttypes',
'django.contrib.sessions',
'django.contrib.messages',
'django.contrib.staticfiles',
# custom apps below
'my_app'
]
Otherwise it will cause an error. 否则会导致错误。 See here: Issue with Django admin registering an inline user profile admin
请参阅此处: Django管理员注册内联用户配置文件admin的问题
To get rid of Users and Groups I had to do in admin.py: 要摆脱我必须在admin.py中执行的用户和组:
from django.contrib import admin
# Need to import this since auth models get registered on import.
import django.contrib.auth.admin
import django.contrib.auth.models
from django.contrib import auth
admin.site.unregister(auth.models.User)
admin.site.unregister(auth.models.Group)
If you got: 如果你有:
django.contrib.admin.sites.NotRegistered: The model Group is not registered
django.contrib.admin.sites.NotRegistered:模型组未注册
Then make sure that your INSTALLED_APPS in proper order like this: 然后确保您的INSTALLED_APPS正确顺序如下:
enter code hereINSTALLED_APPS = (
# [1] Django apps
'django.contrib.auth',
...
# [2] your custom apps
'anyproject.anytuff',
)
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