Problem with Django url regular expression

Question

I am attempting to define a search url using Django with the following structure

www.example.com/afdata/search?device=television&category=news&query=channel&limit=30/

My goal is to extract the values television, news, channel and 30 from the above url and pass it to a view defined as follows

def search(request, device='all', category='single', query='', limit=30):
    return HttpResponse("device=%s, category=%s, query=%s, limit=%d", device, category, query, limit)

afdata is the app and in the url configuration file I defined the url as follows:

from django.conf.urls.defaults import *

urlpatterns = patterns('afdata.views',
    (r'^$', 'index'),
    (r'^search?device=(?P<device>.*)&category=(?P<category>.*)&query=(?P<query>.*)&limit=(?P<limit>d+)/$', 'search')
)

When I run using the above search query in the browser, I get 500 Internal Server Error. index responds fine. Any suggestions on what I may be doing wrong?

James Khoury · Accepted Answer · 2011-06-14 05:45:38Z

5

i suggest that you change your search url to:

(r'^search', 'search')

then in your view do

def search(request):
    device = request.GET.get('device', 'all')
    category = request.GET.get('category', 'single')
    query = request.GET.get('query', '')
    limit = request.GET.get('limit', 30)

   return HttpResponse("device=%s, category=%s, query=%s, limit=%d", device, category, query, limit)

This is much more robust and changeable and flexible.

answered Jun 14, 2011 at 5:45

James Khoury

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3 Comments

David Over a year ago

@James, would the url be structured as follows? www.example.com/search?device=television?category=news?query=channel?limit=30

James Khoury Over a year ago

@david it would be www.example.com/search?device=television&category=news&query=channel&limit=30 as you can't have ? in a url except to separate the url from the parameters

James Khoury Over a year ago

en.wikipedia.org/wiki/Query_string#Structure for more information about query strings/url

Luke Maurer · Accepted Answer · 2011-06-14 05:58:21Z

Django's URL patterns don't match the query part of the URL (the part starting with the ?). You only do what you're trying to do if you have parameters embedded in the URL structure itself.

See the Django docs for the exact details.

So if your URL were:

www.example.com/afdata/search/device/television/category/news/query/channel/limit/30/

Then you would match it with an URL pattern like:

urlpatterns = patterns('afdata.views',
    (r'^$', 'index'),
    (r'^search/device/(?P<device>.*)/category/(?P<category>.*)/query/(?P<query>.*)/limit/(?P<limit>d+)/$', 'search')
)

(This would be a somewhat odd URL convention, but it would work.)

If your URLs have queries in them, you match just the part before the ? and then your view needs to parse out the parameters from request.GET. (If you already have a function like that in your example, it's easy enough to write some glue code to pass the arguments.)

jcomeau_ictx · Accepted Answer · 2011-06-14 05:34:54Z

1

Look in your Apache logs for the stacktrace. also: you sure you don't mean HTTPResponse not HTTPRequest on the 2nd line of your search()?

answered Jun 14, 2011 at 5:34

jcomeau_ictx

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1 Comment

David Over a year ago

Yes, HttpResponse is correct. I tried it but still the same problem. I will look at the stack trace.

Ben Roux · Accepted Answer · 2011-06-14 05:44:46Z

1

By using .* Django might be grabbing up through your &category. Even if this is not the case, it is usually good practice to structure a regex in a way that disallows this. I would consider changing your .* to [^&]*

answered Jun 14, 2011 at 5:44

Ben Roux

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