Django TypeError when passing arguments via URL

Question

Hello I am getting the following error in Django (latest version):

TypeError at /post/1/
post() got an unexpected keyword argument 'post_id'

This occurs when I press on a link on the home page to view the post it's self, I am trying to pass along the post's id (I am using the default [hidden] primary key, not my own custom one)

This is what my urls.py looks like for both the index and post page:

from django.conf.urls.defaults import patterns, include, url
from journal.models import Post


# Uncomment the next two lines to enable the admin:
from django.contrib import admin
admin.autodiscover()

urlpatterns = patterns('journal.views',
    (r'^$', 'index'),
    (r'^post/(?P<id>\d+)/$', 'post'),

Here is my views.py:

from django.http import HttpResponse
from journal.models import Post
from django.template import Context, loader
import os

# Hardcoded Varibles
SITE_ROOT = os.path.join(os.path.dirname(__file__))

# Create your views here.
def index(request):
    latest_post_list = Post.objects.all().order_by('-pub_date')[:10]
    t = loader.get_template(os.path.join(SITE_ROOT, 'templates', 'index.html'))
    c = Context({
    'latest_post_list': latest_post_list,
    })
    return HttpResponse(t.render(c))

def post(request, id):
    return HttpResponse("Hello this is post %" %(post_id))

Nick Craig-Wood · Accepted Answer · 2011-07-17 20:05:13Z

2

Change

def post(request, id):
    return HttpResponse("Hello this is post %" %(post_id))

to

def post(request, id):
    return HttpResponse("Hello this is post %s" % id)

And I suspect it will work a bit better!

answered Jul 17, 2011 at 20:05

Nick Craig-Wood

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2 Comments

jayduff Over a year ago

Ah yes, I had it as "id" before, however I have been missing the s from %s that part, thank you.

Nick Craig-Wood Over a year ago

Looks like you forgot the %s on the end? In your original you forgot it, so maybe you didn't change it when you tried the above?

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