Question

我想要匹配的网址是：

此请求是通过ajax从外部api获取一些JSON数据。我不能得到它来匹配;继续我已经设置的中间件处理程序。我确定这个修复是非常明显的，但我一直盯着这个方式太久了。

urls.py：

from django.conf import settings
from django.conf.urls.defaults import patterns, include
from django.contrib import admin


admin.autodiscover()

urlpatterns = patterns('',

    (r'^$', 'blog.views.index'),
    (r'^games/$', 'giantbomb_games.views.index'),
    (r'^games/getdata/(?P<resource>\w+)/$', 'giantbomb_games.views.getdata'),

    (r'^grappelli/', include('grappelli.urls')),
    (r'^admin/filebrowser/', include('filebrowser.urls')),
    (r'^admin/', include(admin.site.urls)),
)

views.py：

def getdata(request, resource):

    url = '%s/%s/?api_key=%s&format=%s' % (api_url, resource , api_key, request_format)
    print url
    r = requests.get(url)

    return r.content

页/ middleware.py：

from django.http import Http404
from django.conf import settings

from page.views import site_page

class SitePageFallbackMiddleware(object):
    def process_response(self, request, response):
        if response.status_code != 404:
            return response # No need to check for a flatpage for non-404 responses.
        try:
            return site_page(request, request.path_info)
        # Return the original response if any errors happened. Because this
        # is a middleware, we can't assume the errors will be caught elsewhere.
        except Http404:
            return response
        except:
            if settings.DEBUG:
                raise
            return response

django错误：

Traceback (most recent call last):

  File "/var/www/html/top10/top10/../ext/django/core/servers/basehttp.py", line 280, in run
    self.result = application(self.environ, self.start_response)

  File "/var/www/html/top10/top10/../ext/django/core/servers/basehttp.py", line 674, in __call__
    return self.application(environ, start_response)

  File "/var/www/html/top10/top10/../ext/django/core/handlers/wsgi.py", line 245, in __call__
    response = middleware_method(request, response)

  File "/var/www/html/top10/top10/page/middleware.py", line 8, in process_response
    if response.status_code != 404:

AttributeError: 'str' object has no attribute 'status_code'

提前谢谢。

Answer 1

我不确定我是否理解正确...你是说你的观点没有被称呼？

如果您已经设置了一些中间件类，那么处理请求的流在之前（和之后）通过中间件是正常的，它会到达您的视图。

以下是一些文档：

https://docs.djangoproject.com/en/dev/topics/http/middleware/?from=olddocs

在你的中间件类中，当它到达方法时，你应该能够看到url解析器将调用哪个视图函数： process_view（self，request，view_func，view_args，view_kwargs）

如果我不太明白你的问题，请告诉我。

Answer 2

getdata视图返回字符串作为响应而不是HttpResponse对象，因此中间件中的process_response获取字符串而字符串没有status_code方法。将您的观点更改为：

def getdata(request, resource):
    #...
    return HttpResponse(r.content, mimetype="text/plain")

Answer 3

想出来。 Django实际上正确地击中了视图。由于在getdata视图中发出了格式错误的url请求，因此api抛出了一个错误。我非常愚蠢。

django应用程序视图的urls.py中的正则表达式匹配

3 个答案: