Django - Dynamic view for url

Question

I want to load a particular view depending on the url, for example:

url(r'^channel/(?P<channel>\d+)/$', ---, name='channel_render'),

Depending on the channel passed into the url, I want to load a specific view file. I tried doing this:

def configure_view(channel):
    print channel

urlpatterns = patterns('',
    url(r'^channel/(?P<channel>\d+)/$', configure_view(channel), name='channel_render'),

But obviously the channel argument is not getting passed in. Is there any way to do this? The only other solution I can think of is loading a manager view and then loading the relevant view file from there. If this is the only way, how do I redirect to another view file from within a view?

tino · Accepted Answer · 2013-11-12 15:40:21Z

4

You could do something like this.

#urls.py
url(r'^channel/(?P<channel>\d+)/$', switcher, name='channel_render'),

#views.py
def switcher(request, channel):
    if channel == 'Whatever':
        return view_for_this_channel()

def view_for_this_channel()
    #handle like a regular view

If using class-based views, the call in your switcher() will look like this:

#views.py
def switcher(request, channel):
    if channel == 'Whatever':
        return ViewForThisChannel.as_view()(request)  # <-- call to CBV

def ViewForThisChannel(View):
    #handle like a regular class-based view

edited Nov 12, 2013 at 15:40

tino

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answered May 20, 2011 at 14:54

silent1mezzo

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Comments

phedoreanu · Accepted Answer · 2014-08-16 21:26:56Z

2

For redirecting you should use the Django redirect shortcut function:

from django.shortcuts import redirect

def my_view(request):
    ...
    return redirect('some-view-name', foo='bar')

https://docs.djangoproject.com/en/1.7/topics/http/shortcuts/#redirect

edited Aug 16, 2014 at 21:26

answered Aug 16, 2014 at 21:18

phedoreanu

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Comments

Luke Sneeringer · Accepted Answer · 2011-05-20 14:53:05Z

1

I think the easiest way to do this is to load a view that functions as a tiny dispatcher, which calls the final view you're interested in.

As far as how to do that, views are just functions that get called in a particular way and expected to return a particular thing. You can call one view from another; just make sure you're properly returning the result.

You can load views from different files with import.

answered May 20, 2011 at 14:53

Luke Sneeringer

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2 Comments

Hanpan Over a year ago

I'd rather load the view in the same way django does when the URL is resolved, just cold importing the view doesn't seem right? I'll have a look through the source code and see how django imports views, perhaps there is a function I can hijack.

Hanpan Over a year ago

Seems I could use import_module from django.utils.importlib. Thanks.

JamesO · Accepted Answer · 2011-05-20 14:54:15Z

0

try calling like a normal view e.g.

def configure_view(request, channel):
    print channel

url(r'^channel/(?P<channel>\d+)/$', configure_view, name='channel_render'),

answered May 20, 2011 at 14:54

JamesO

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