Django:无法访问中间件中的变量集

时间:2017-10-03 23:10:55

标签: django django-middleware

我正在尝试在我的自定义中间件中向我的请求变量添加变量。后来我想从我的视图中访问该变量。以下是相关内容:

middleware.py

class DomainInterceptMiddleware(object):
    def process_request(self, request):
        request.myvar = 'whatever'

settings.py - 上面添加了CommonMiddleware

MIDDLEWARE = [
    'django.middleware.security.SecurityMiddleware',
    'django.contrib.sessions.middleware.SessionMiddleware',
    'page.middleware.DomainInterceptMiddleware',  #
    'django.middleware.common.CommonMiddleware',
    'django.middleware.csrf.CsrfViewMiddleware',
    'django.contrib.auth.middleware.AuthenticationMiddleware',
    'django.contrib.messages.middleware.MessageMiddleware',
    'django.middleware.clickjacking.XFrameOptionsMiddleware',
]

views.py

def index(request):
    output = request.myvar
    return HttpResponse(output)

但是,在尝试访问/index时,我收到此错误

  File "/path/to/views.py", line 32, in index
    output = request.myvar
AttributeError: 'WSGIRequest' object has no attribute 'myvar'

我哪里错了?

1 个答案:

答案 0 :(得分:3)

您当前的中间件类是针对旧MIDDLEWARE_CLASSES设置编写的。

您正在使用MIDDLEWARE设置,因此您需要编写new-style middleware(Django 1.10+)。

如果您需要向后兼容,可以使用django.utils.deprecation中的MiddlewareMixin

from django.utils.deprecation.MiddlewareMixin

class DomainInterceptMiddleware(MiddlewareMixin):
    def process_request(self, request):
        request.myvar = 'whatever'

或者您可以将其转换为新式中间件。

class DomainInterceptMiddleware(object):
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        request.myvar = 'whatever'
        response = self.get_response(request)
        return response

有关详细信息,请参阅writing your own middleware上的文档。