“ WSGIRequest”对象没有属性“ update”

Question

我已经针对类似的问题通过了一些解决方案，但是这些都不起作用：

“ WSGIRequest”对象没有属性“ session”

“ WSGIRequest”对象没有属性“ facebook”

我正在建立一个搜索页面，如下所示：

views.py

def search(request):
    if 'query' in request.GET and request.GET['query']:
        q = request.GET['query']
        ct = Product.objects.filter(Q(name__icontains=q) | Q(desc__icontains=q) | Q(category__icontains=q))
        count = ct
        paginate = Paginator(ct, 10)
        page = request.GET.get('page')
        try:
            ct = paginate.page(page)
        except PageNotAnInteger:
            ct = paginate.page(1)
        except EmptyPage:
            ct = paginate.page(paginate.num_pages)
        dictionary = {'results': ct, 'count': count, 'query': q, }
        return render_to_response('search.html', dictionary, request)
    else:
        content = {'contentnotfound': 'Hi, you did not search for anything. Please go back or use the search box above.'}
        return render(request, 'search.html', content)

完整的回溯是：

Environment:


Request Method: GET
Request URL: http://127.0.0.1:8000/search?query=abc&submit=Submit

Django Version: 1.6.4
Python Version: 2.7.3
Installed Applications:
('django_admin_bootstrapped.bootstrap3',
 'django_admin_bootstrapped',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'Abstract',
 'south',
 'djrill',
 'storages',
 's3direct')
Installed Middleware:
('django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware')


Traceback:
File "/home/siddharth/AbstractIndia/venv/local/lib/python2.7/site-packages/django/core/handlers/base.py" in get_response
  114.                     response = wrapped_callback(request, *callback_args, **callback_kwargs)
File "/home/siddharth/AbstractIndia/Abstract/views.py" in search
  39.         return render_to_response('search.html', dictionary, request)
File "/home/siddharth/AbstractIndia/venv/local/lib/python2.7/site-packages/django/shortcuts/__init__.py" in render_to_response
  29.     return HttpResponse(loader.render_to_string(*args, **kwargs), **httpresponse_kwargs)
File "/home/siddharth/AbstractIndia/venv/local/lib/python2.7/site-packages/django/template/loader.py" in render_to_string
  167.     context_instance.update(dictionary)

Exception Type: AttributeError at /search
Exception Value: 'WSGIRequest' object has no attribute 'update'

urls.py如下：

 url(r'^search$', views.search, name='search'),

在另一个项目之前，我已经做过同样的事情，并且效果很好。

Answer 1

return render_to_response('search.html', dictionary, request)

您将请求作为上下文实例传递，而应该传递django.template.Context实例（尽管您可能希望使用其子类RequestContext ）。

Answer 2

您已经在最后一行正确使用了新的render快捷方式。 但是，在else子句之前的行中，您使用了较旧的render_to_response并将请求作为最后一个参数传递。 您也应该在此处使用render ，并将request作为第一个参数：

return render(request, 'search.html', dictionary)