django get complete url in query string

Question

I am trying to implement a redirection script. The format of the url would be

http://localhost:8000/key/url=http://google.com

From the above, I want http://google.com

When some user visits the above url, it hits the urlpatters defined in the urls.py

url(r'^key/url=(.*)', 'homepage.views.redirectquerystring', name="Redirect"),

I am trying to get the url http://google.com using the below view

def redirectquerystring(request):
    para = request.GET.get('url','')

But when I do this, I am getting the following error:

TypeError at /key/url=http://google.com
redirectquerystring() takes exactly 1 argument (2 given)

Am I doing some mistake here.

Thanks.

Daniel Rosenthal · Accepted Answer · 2013-08-12 15:28:16Z

3

This is much simpler than you think it is.

You're trying to pass http://google.com as a parameter, but you're not giving your view a place to receive that parameter.

You need to define your view as def redirectqyrystring(request, url):

You don't need to get the url from the request now, it's already there in the variable url

answered Aug 12, 2013 at 15:28

Daniel Rosenthal

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Comments

zaadeh · Accepted Answer · 2013-08-12 15:31:58Z

3

You should "capture" the url param using the urlpatterns regex like this:

url(r'^key/url=(?P<url>.*)', 'homepage.views.redirectquerystring', name="Redirect"),

this way your view receives a parmeter named url, which contains the captured url get param.

answered Aug 12, 2013 at 15:31

zaadeh

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def redirectquerystring(request, url):

django is beautiful, isn't it?