Python django admin，使用自定义视图扩展admin

我想就这件事请求一些帮助

我已经按照指南为我的管理员添加了一个视图

我使用的代码与网站的代码相同，但出现了一个错误：

找不到参数为“（）”且关键字参数为“{'app_label'：'reports'}”的“app_list”的反转。

admin/my_视图

工作正常，但是如果我试图访问管理中的其他页面，例如索引页面，我会收到错误消息

代码如下：

#someapp/admin.py

from django.conf.urls import patterns
from django.contrib import admin
from django.http import HttpResponse

def my_view(request):
    return HttpResponse("Hello!")

def get_admin_urls(urls):
    def get_urls():
        my_urls = patterns('',
            (r'^my_view/$', admin.site.admin_view(my_view))
         )
        return my_urls + urls
    return get_urls

admin_urls = get_admin_urls(admin.site.get_urls())
admin.site.get_urls = admin_urls

我使用的是django 1.8.2和python 2.7

另外，在管理员的索引页面中添加一些视图或上下文的最佳方式是什么

更新

我试过安托万·平萨德的答案，并试过：

#admin.py
from django.contrib.admin import AdminSite
class MyAdminSite(AdminSite):

    def get_urls(self):
        from django.conf.urls import url
        urls = super(MyAdminSite, self).get_urls()
        urls += [
            url(r'^my_view/$', self.admin_view(self.my_view))
        ]
        return urls

    def my_view(self, request):
        return HttpResponse("Hello!")

admin_site = MyAdminSite()

并将url.py的

url（r'^admin/，include（admin.site.url）），

替换为

url（r'^admin/，include（admin\u site.url）），

并获得：

带参数“（）”和关键字参数的“app_list”反向

回溯：

Request Method: GET
Request URL: http://localhost:8000/@dmin/

Django Version: 1.8.2
Python Version: 2.7.10
Installed Applications:
('grappelli',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'django.contrib.sessions.models',
 'frontend',
 'file_maintenance',
 'reports',
 'transactions',
 'admin_reorder',
 'admin_notifications',
 'django_twilio',
 'daterange_filter')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
 'django.contrib.auth.models.User',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware',
 'django.middleware.security.SecurityMiddleware',
 'admin_reorder.middleware.ModelAdminReorder')


Traceback:
File "C:\Python27\lib\site-packages\django\core\handlers\base.py" in get_response
  157.                     response = middleware_method(request, response)
File "C:\Python27\lib\site-packages\admin_reorder\middleware.py" in process_template_response
  134.         self.init_config(request, app_list)
File "C:\Python27\lib\site-packages\admin_reorder\middleware.py" in init_config
  21.         admin_index = admin.site.index(request)
File "C:\Python27\lib\site-packages\django\views\decorators\cache.py" in _wrapped_view_func
  57.         response = view_func(request, *args, **kwargs)
File "C:\Python27\lib\site-packages\django\contrib\admin\sites.py" in index
  438.                                 current_app=self.name,
File "C:\Python27\lib\site-packages\django\core\urlresolvers.py" in reverse
  579.     return force_text(iri_to_uri(resolver._reverse_with_prefix(view, prefix, *args, **kwargs)))
File "C:\Python27\lib\site-packages\django\core\urlresolvers.py" in _reverse_with_prefix
  496.                              (lookup_view_s, args, kwargs, len(patterns), patterns))

Exception Type: NoReverseMatch at /@dmin/
Exception Value: Reverse for 'app_list' with arguments '()' and keyword arguments '{'app_label': 'django_twilio'}' not found. 0 pattern(s) tried: []

这本指南看起来很旧。我宁愿建议你照办

someapp/admin.py

from django.contrib.admin import AdminSite
from django.http import HttpResponse

class MyAdminSite(AdminSite):

     def get_urls(self):
         from django.urls import path
         urls = super().get_urls()
         urls += [
             path('my_view/', self.admin_view(self.my_view))
         ]
         return urls

     def my_view(self, request):
         return HttpResponse("Hello!")

admin_site = MyAdminSite()

资料来源：

您还应该更新您的
项目/url.py
并将
路径（'admin/'，admin.site.url）
替换为
路径（'admin/'，admin\u site.url）
。不要忘记以前从someapp.admin导入admin_站点
。
您链接的指南很旧，我很惊讶在过去一年左右没有找到任何直接回答您问题的内容

在应用程序的admin.py中或在方便的地方创建一个新应用程序
在新的AdminSite中创建一个函数，用您的额外URL扩展
get\u url（）
函数
确保您的项目URL.py链接到新创建的AdminSite
以下代码适用于Python 3.5.1和Django 1.9.6


我的应用程序/admin.py

from django.contrib import admin
from django.contrib.admin import AdminSite
from django.http import HttpResponse

from my_app.models import SomeModel


class MyAdminSite(AdminSite):

    def custom_view(self, request):
        return HttpResponse("Test")

    def get_urls(self):
        from django.conf.urls import url
        urls = super(MyAdminSite, self).get_urls()
        urls += [
            url(r'^custom_view/$', self.admin_view(self.custom_view))
        ]
        return urls

admin_site = MyAdminSite()


@admin.register(SomeModel, site=admin_site)
class SomeModelAdmin(admin.ModelAdmin):
    pass

from django.conf.urls import url, include

from my_app.admin import admin_site

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

myu项目/url.py

from django.contrib import admin
from django.contrib.admin import AdminSite
from django.http import HttpResponse

from my_app.models import SomeModel


class MyAdminSite(AdminSite):

    def custom_view(self, request):
        return HttpResponse("Test")

    def get_urls(self):
        from django.conf.urls import url
        urls = super(MyAdminSite, self).get_urls()
        urls += [
            url(r'^custom_view/$', self.admin_view(self.custom_view))
        ]
        return urls

admin_site = MyAdminSite()


@admin.register(SomeModel, site=admin_site)
class SomeModelAdmin(admin.ModelAdmin):
    pass

from django.conf.urls import url, include

from my_app.admin import admin_site

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

对于Django 1.4+，以下是解决方案：

from django.conf.urls import url
from django.contrib import admin
from .models import MyModel


class MyAdmin(admin.ModelAdmin):
    list_display = (...)

    def custom_admin_view(self, request):
        # your logic here
        if request.method == 'POST':
            ...
        else:
            ...
        return HttpResponse(...)

    def get_urls(self):
        additional_urls = [
            url(r'^custom/$', self.admin_site.admin_view(self.custom_admin_view), name='custom')
        ]
        # append your custom URL BEFORE default ones
        return additional_urls + super().get_urls()


admin.site.register(MyModel, MyAdmin)

在所有其他URL之前添加自定义URL非常重要。否则，您可能会被重定向到admin
change\u view
 下面的代码将作为默认的管理员站点来编写自定义视图

适用于Django>=2.1：

###myproject/admin.py

from django.contrib import admin
from django.http import HttpResponse

class MyAdminSite(admin.AdminSite):

    def my_view(self, request): # your custom view function
        return HttpResponse("Test")

    def get_urls(self):
        from django.urls import path

        urlpatterns = super().get_urls()
        urlpatterns += [
            path('my_view/', self.admin_view(self.my_view))
        ]
        return urlpatterns

###myproject/apps.py

from django.contrib.admin.apps import AdminConfig

class MyAdminConfig(AdminConfig):
    default_site = 'myproject.admin.MyAdminSite'

###myproject/settings.py

INSTALLED_APPS = [
    ...
    'myproject.apps.MyAdminConfig',  # replaces 'django.contrib.admin'
    ...
]

然后转到
请提供完整的回溯。另外，请注意，模式已被弃用。你应该使用
url̀
列表。我已经更新了我的帖子，如果还有人对此有问题，你也可以查看该帖子。这有点棘手，但有助于避免在项目上覆盖AdminSite实例。我在Upvoted上详细介绍了如何扩展admin，以获得一个清晰有效的示例。你知道如何在多个项目而不是一个项目中实现这一点吗？@dKen我很困惑。你是说多个Django项目吗？如果是这样的话，我不确定这是否可能。非常抱歉，我把我的命名法弄错了！我想问的是如何让它与多个应用程序一起工作。抱歉搞混了！嗯，这将适用于多个应用程序。。。您只需将实例化的类导入到您的所有应用程序中：
从common.admin导入admin_站点
@DanielvanFlymen，第三方应用程序admin是否会以这种方式包含工作？我的意思是，一些软件包在安装并添加到已安装的应用程序之后，自动将自己绑定到索引页，并具有所有功能。