django and url creation

Question

I want to construct the following url

domain/edit/xray/<id>/<image_name>

where id is a model's id and image_name is the name of an image.

I have an app that will handle image editing with its own views lets say it mypil

mypil.views

def edit_xray(request, id, image_name):
    ....code to get image by name
    ....model by id and edit image using PIL
        model = Model.objects.get(pk=id)

project/urls.py

url(r'^edit/', include(mypil.urls))

mypil.urls

url(r'^xray/(?P<id>)\d+/(?P<image_name>)\w+\.\w{3}$', mypil.views.edit_xray)

But when i try to view edit/xray/1/image.jpg (both id and image exist) i encounter this problem.

invalid literal for int() with base 10: ''

and the traceback shows me that line above

model=Model.objects.get(pk=id)

Does the empty string('') mean that it doesn't parse the id correctly from the URL? Isn't my url pattern correct?

EDIT: Damn my eyes....needed to put \d+ and the image reg pattern inside the parentheses

url(r'^xray/(?P<id>\d+)/(?P<image_name>\w+\.\w{3})$,...)

Sorry for posting a question...(How do i delete my own questions?)

Don't delete it, it's not a bad question. Someone might do the same mistake again — yuvi
– yuvi, Commented Oct 29, 2013 at 10:41

iMom0 · Accepted Answer · 2013-10-29 18:39:26Z

1

The correct url is:

url(r'^xray/(?P<id>\d+)/(?P<image_name\w+\.\w{3})$', mypil.views.edit_xray)

You should put the pattern in the parentnesses.

edited Oct 29, 2013 at 18:39

answered Oct 29, 2013 at 10:40

iMom0

12.9k3 gold badges52 silver badges61 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

iMom0 Over a year ago

Keeping the trailing slash is a good practice, in my opinion.

Collectives™ on Stack Overflow

django and url creation

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related