Can't fetch offset value in the URL （Django）

Question

Possible Duplicate:
Capturing url parameters in request.GET

I'm designing a detail page of my website by using django, and set the Topic Url as following:

(r'd/\d+/$', 'xiangwww.detail.views.detail'),

the second argument is the ID of the topic. But when I fetch it following the guide of Django Book :

def detail(request,offset):
    print offset
    return render_to_response('detail.html')

The page "localhost:8000/d/1/" in shows the TypeError: detail() takes exactly 2 arguments (1 given)

It seems Django can't recognise what is offset in my views.py file, how to solve it?

Martijn Pieters · Accepted Answer · 2018-09-28T11:23:08.110

4

You need to capture the number, by putting it in a regular expression group:

(r'd/(\d+)/$', 'xiangwww.detail.views.detail'),

Without the (...) group, Django does not know about the captured number and cannot pass it to your view. See the URL Dispatch documentation:

To capture a value from the URL, just put parenthesis around it.

edited Sep 28 '18 at 11:23

answered Sep 24 '12 at 11:26

Martijn Pieters

is there any possibility to pass more than two arguments as offsets? – Wei Lin Sep 24 '12 at 14:59
Sure, just create more `(..)` groups; the linked documentation has some examples that use multiple capturing groups. – Martijn Pieters Sep 24 '12 at 15:07

score 0 · Answer 2 · answered Sep 24 '12 at 11:26

0

You need to change url to accept offset as parameter e.g.

(r'd/(?P<offset>\d+/)$', 'xiangwww.detail.views.detail'),

answered Sep 24 '12 at 11:26

Rohan

2 Answers2