Question

在Django中有一个设置文件，用于定义要在每个请求上运行的中间件。此中间件设置是全局的。有没有办法在每个视图的基础上指定一组中间件？我希望特定的URL使用一组与全局集不同的中间件。

Answer 1

你想要decorator_from_middleware。

from django.utils.decorators import decorator_from_middleware

@decorator_from_middleware(MyMiddleware)
def view_function(request):
    #blah blah

它不适用于网址，但它适用于每个视图，因此您可以对其效果进行细粒度控制。

Answer 2

我对这个问题有一个真正的解决方案。警告;这有点像黑客。

""" Allows short-curcuiting of ALL remaining middleware by attaching the
@shortcircuitmiddleware decorator as the TOP LEVEL decorator of a view.

Example settings.py:

MIDDLEWARE_CLASSES = (
    'django.middleware.common.CommonMiddleware',
    'django.contrib.sessions.middleware.SessionMiddleware',
    'django.middleware.csrf.CsrfViewMiddleware',
    'django.contrib.auth.middleware.AuthenticationMiddleware',
    'django.contrib.messages.middleware.MessageMiddleware',

    # THIS MIDDLEWARE
    'myapp.middleware.shortcircuit.ShortCircuitMiddleware',

    # SOME OTHER MIDDLE WARE YOU WANT TO SKIP SOMETIMES
    'myapp.middleware.package.MostOfTheTimeMiddleware',

    # MORE MIDDLEWARE YOU WANT TO SKIP SOMETIMES HERE
)

Example view to exclude from MostOfTheTimeMiddleware (and any subsequent):

@shortcircuitmiddleware
def myview(request):
    ...

"""

def shortcircuitmiddleware(f):
    """ view decorator, the sole purpose to is 'rename' the function
    '_shortcircuitmiddleware' """
    def _shortcircuitmiddleware(*args, **kwargs):
        return f(*args, **kwargs)
    return _shortcircuitmiddleware

class ShortCircuitMiddleware(object):
    """ Middleware; looks for a view function named '_shortcircuitmiddleware'
    and short-circuits. Relies on the fact that if you return an HttpResponse
    from a view, it will short-circuit other middleware, see:
    https://docs.djangoproject.com/en/dev/topics/http/middleware/#process-request
     """
    def process_view(self, request, view_func, view_args, view_kwargs):
        if view_func.func_name == "_shortcircuitmiddleware":
            return view_func(request, *view_args, **view_kwargs)
        return None

编辑：删除了两次运行视图的先前版本。

Answer 3

这是我最近用来解决你在对Ned的回答的评论中提出的场景的解决方案......

它假设：

A）这是一个自定义中间件或者您可以使用自己的中间件类扩展/包装的中间件

B）您的逻辑可以等到process_view而不是process_request，因为在process_view中，您可以在解析后检查view_func参数。（或者您可以调整下面的代码以使用Ignacio指示的urlresolvers。

# settings.py
EXCLUDE_FROM_MY_MIDDLEWARE = set('myapp.views.view_to_exclude', 
    'myapp.views.another_view_to_exclude')

# some_middleware.py

from django.conf import settings

def process_view(self, request, view_func, view_args, view_kwargs):
    # Get the view name as a string
    view_name = '.'.join((view_func.__module__, view_func.__name__))

    # If the view name is in our exclusion list, exit early
    exclusion_set = getattr(settings, 'EXCLUDE_FROM_MY_MIDDLEWARE', set())
    if view_name in exclusion_set:
        return None

    # ... middleware as normal ...
    #
    # Here you can also set a flag of some sort on the `request` object
    # if you need to conditionally handle `process_response` as well.

可能有一种方法可以进一步推广这种模式，但这很好地实现了我的目标。

为了回答您更普遍的问题，我认为Django库中没有任何东西可以帮助您解决这个问题。对于django-users邮件列表来说，如果它还没有在那里解决，那将是一个很好的主题。

Answer 4

您可以使用在调用view func之前调用的process_view方法。在process_view中，您可以检查 - 此视图是否需要此中间件拦截。

Answer 5

在中间件的包装器中对request.path使用django.core.urlresolvers.resolve()以尝试查看视图是否在应用内，如果是，则跳过处理。

Answer 6

我能找到的最好的事情是使用if request.path_info.startswith（'...'）通过返回请求来跳过中间件。现在，您可以仅为了跳过而创建中间件，然后继承它。也许你可以做一些更简单的事情，并在settings.py中保存该列表，然后跳过所有这些。如果我错了，请告诉我。

Answer 7

Django urlmiddleware允许仅将中间件应用于映射到特定网址的视图。

Answer 8

我认为这是从中间件中排除视图的简单方法

 from django.core.urlresolvers import resolve
 current_url = resolve(request.path_info).url_name

 if want to exclude url A,

 class your_middleware:
    def process_request(request):
        if not current_url == 'A':
            "here add your code"

Answer 9

#settings.py
EXCLUDE_FROM_MY_MIDDLEWARE =set({'custom_app.views.About'})

#middlware.py
from django.conf import settings

class SimpleMiddleware(object):

     def __init__(self,get_response):
          self.get_response=get_response
          
     
     def __call__(self,request):
          
          response = self.get_response(request)
          return response

     def process_view(self,request, view_func, view_args, view_kwargs):
         
          view_function='.'.join((view_func.__module__,view_func.__name__))
          exclusion_set=getattr(settings,'EXCLUDE_FROM_MY_MIDDLEWARE',set() )
          if view_function in exclusion_set:
               return None
          
          print("continue for others views")
          
     def process_exception(self,request, exception):
          return HttpResponse(exception)

Django中的非全局中间件

9 个答案: