What do you mean,
If i take it out of the ctx dictionary how do I pass it. if I just declare
the @csrf_protect I still get the token error trying to post my form,
infact the csrf issue is the only thing I have left to fix. How are you
supposed to do the csrf token because im following the documentation and
its not working. I tried what you suggested removing the declaration no
luck.
On Thursday, May 15, 2014 8:12:33 AM UTC-6, donarb wrote:
>
> On Wednesday, May 14, 2014 11:35:46 PM UTC-7, G Z wrote:
>>
>> So I read the documentation on passing csrf tokens, however its giving me
>> an issue i think its because im trying to pass it as a dictonary variable
>> with my form and customers.
>>
>> This is from the documentation
>>
>> from django.views.decorators.csrf import csrf_protectfrom django.shortcuts
>> import render
>> @csrf_protectdef my_view(request):
>> c = {}
>> # ...
>> return render(request, "a_template.html", c)
>>
>>
>>
>> from django.shortcuts import render
>> from django.http import HttpResponse
>> from vmware.models import Customer
>> from django.shortcuts import render_to_response
>> from vmware.models import Vms
>> from .forms import SignUpForm
>> from vmware.models import Vmspecs
>> from django.views.decorators.csrf import csrf_protect
>>
>> @csrf_protect
>> def index(request):
>> c = {}
>> form = SignUpForm(request.POST or None)
>> if form.is_valid():
>> save_it = form.save(commit=False)
>> save_it.save()
>> customers = Customer.objects.all()
>> ctx = { 'customers':customers, 'form':form, c}
>> return render_to_response('index.html', ctx)
>>
>>
>>
>> this is the error i get
>>
>>
>> Environment:
>>
>>
>> Request Method: GET
>> Request URL: http://23.239.206.142:8000/
>>
>> Django Version: 1.6.4
>> Python Version: 2.7.3
>> Installed Applications:
>> ('django.contrib.admin',
>> 'django.contrib.auth',
>> 'django.contrib.contenttypes',
>> 'django.contrib.sessions',
>> 'django.contrib.messages',
>> 'django.contrib.staticfiles',
>> 'vmware')
>> 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
>> "/usr/local/lib/python2.7/dist-packages/django/core/handlers/base.py" in
>> get_response
>> 101. resolver_match =
>> resolver.resolve(request.path_info)
>> File "/usr/local/lib/python2.7/dist-packages/django/core/urlresolvers.py"
>> in resolve
>> 337. for pattern in self.url_patterns:
>> File "/usr/local/lib/python2.7/dist-packages/django/core/urlresolvers.py"
>> in url_patterns
>> 365. patterns = getattr(self.urlconf_module, "urlpatterns",
>> self.urlconf_module)
>> File "/usr/local/lib/python2.7/dist-packages/django/core/urlresolvers.py"
>> in urlconf_module
>> 360. self._urlconf_module = import_module(self.urlconf_name)
>> File "/usr/local/lib/python2.7/dist-packages/django/utils/importlib.py"
>> in import_module
>> 40. __import__(name)
>> File "/root/djangoprojects/provisioning/provisioning/urls.py" in <module>
>> 7. url(r'^customers/', include('vmware.urls')),
>> File
>> "/usr/local/lib/python2.7/dist-packages/django/conf/urls/__init__.py" in
>> include
>> 26. urlconf_module = import_module(urlconf_module)
>> File "/usr/local/lib/python2.7/dist-packages/django/utils/importlib.py"
>> in import_module
>> 40. __import__(name)
>> File "/root/djangoprojects/provisioning/vmware/urls.py" in <module>
>> 5. from vmware import views
>>
>> Exception Type: SyntaxError at /
>> Exception Value: invalid syntax (views.py, line 19)
>>
>>
>>
> Specifically, you're passing 'c' into the ctx dictionary, but it has no
> key, that's the syntax error. But that c dictionary is redundant anyway
> since you've created your own context dictionary. Just remove the
> declaration of that object and don't pass it into the dictionary.
>
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