Force django querystring in url

Question

Say you have a url like this:

/cats/?filter=kittens

Is it possible to build a django url pattern that forces the use of the querystring?

Currently I have:

url(r'^/cats/$', views.CatsListView.as_view(), name='cats')

Now I want to add the querystring and get a different view, something like this:

url(r'^/cats/?filter=(?P<filter>.+?)$', views.CatsFilteredListView.as_view(), name='cats-filtered')

Is it possible to do something like this and still keep the querystring in the GET parameter of the request?
Remember that this is just a testcase, I, and you should too, know that filtering like probably this isn't the way to go..

bruno desthuilliers · Accepted Answer · 2018-05-16 12:52:36Z

3

Short answer: no, it's not possible. Django url patterns match only on the "path" componant of the url, period.

answered May 16, 2018 at 12:52

bruno desthuilliers

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Nebulosar Over a year ago

Alright, that clears things up, I was searching the web all over for this :)

Alasdair · Accepted Answer · 2018-05-16 12:54:19Z

1

No, it's not possible to do this. If you really need two separate views, you can write a view that dispatches the correct view.

def cat_list_view(request, *args, **kwargs):
    if 'filter' in request.GET:
        return cat_list_filter_view(request, *args, **kwargs)
    else:
        return cat_list_unfiltered_view(request, *args, **kwargs)

However, for your example of CatsListView and CatsFilteredListView there is probably a better way to combine the views. For example you might be able to do the filtering in the get_queryset method.

answered May 16, 2018 at 12:54

Alasdair

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Nebulosar Over a year ago

Yeah, I was thinking of just checking the param. (And then also keep things inside one view. I agree about the get_queryset Django really has some powerful things to do this. Tnx for your answer!

dbinott · Accepted Answer · 2018-10-31 19:59:50Z

While it is true that you cannot manipulate the path component like that, you can pass a dictionary in. It's a 3rd unnamed argument.

This approach can be useful if you want to use the same view for multiple resources, and pass data to configure its behaviour in each case (below we supply a different template in each case).

path('url/', views.my_reused_view, {'my_template_name': 'some_path'}, name='aurl'),
path('anotherurl/', views.my_reused_view, {'my_template_name': 'another_path'}, name='anotherurl'),

Note: Both extra options and named captured patterns are passed to the view as named arguments. If you use the same name for both a captured pattern and an extra option then only the captured pattern value will be sent to the view (the value specified in the additional option will be dropped).

Courtesy of https://developer.mozilla.org/en-US/docs/Learn/Server-side/Django/Generic_views

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