CODEWITHSUNDEEP
pythondjango
user3114321
user3114321

Reputation: 61

Unable to configure views, urls.py SyntaxError: invalid syntax

I'm having issue while configuring views and I am following django 1.5 official tutorial. Here is my Code for polls/urls.py.

from django.conf.urls import patterns, url

from polls import views

urlpatterns = patterns ('', 
url(r'^$', views.index, name='index')
url(r'^(?P<poll_id>\d+)/$', views.detail, name='detail'), 
url(r'^(?P<poll_id>\d+)/results/$', views.results, name='reults'),
url(r'^(?P<poll_id>\d+/vote/$', views.vote, name='vote'),

)

Below is my polls/views.py

 from django.http import HttpResponse

def index(request):
return HttpResponse("Hello, world. You're at the poll index.")

def detail(request, poll_id):
return HttpResponse("You’re looking at poll %s." % poll_id)

def results(request, poll_id):
return HttpResponse("You’re looking at the results of poll %s." % poll_id)

def vote(request, poll_id):
return HttpResponse("You’re voting on poll %s." % poll_id)

In polls/urls.py I've also tried url(r'^(?P\d+)/detail/$', views.detail, name='detail'), instead of url(r'^(?P\d+)/$', views.detail, name='detail'), Error I'm getting is

File "C:\Python27\Scripts\mysite\polls\urls.py", line 7 url(r'^(?P\d+)/$', views.detail, name='detail'), ^ SyntaxError: invalid syntax [31/Dec/2013 06:06:34] "GET /admin/ HTTP/1.1" 500 84890

Please help.

Upvotes: 1

Views: 3564

Answers (1)

iMom0
iMom0

Reputation: 12931

In your code

url(r'^(?P<poll_id>\d+/vote/$', views.vote, name='vote'),

should be

url(r'^(?P<poll_id>\d+)/vote/$', views.vote, name='vote'),

You missed the parentness and you missed a comma after url(r'^$', views.index, name='index').

Upvotes: 2

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