How to remove get parameter from url in django view

Question

I have the following view:

def copy_group(request):
    copy = request.GET.get('copy','')

    if copy:
        #do my copy process

    context = {'view':'copy-view'}
    return render(request, 'groups/copy-view.html', context)

This invoked with the following url mysite.com/groups/?copy=1

The problem is that if I refresh the page, my process keeps copying over and over.

How can I remove the ge parameter so the url returns after it copies: mysite.com/groups/

I tried inserting this in my view code:

request.GET.pop('copy')

But I get the error: This QueryDict instance is immutable

I think you are doing something wrong, if the request modify data on server, it must be a POST, not a GET — Leandro, Oct 01 '13 at 01:28

score 2 · Answer 1 · answered Oct 01 '13 at 01:30

If I understand it well, once you've done the copy you want the page to be redirected so no more copy will be made. Do like this then:

...
if copy:
    the actual copy and then...
    return HttpResponseRedirect(redirect_to='/the-path-without-copy-parameter/')
...

Then you can improve your code:

add copy as an actual route parameter, def copy_group(request, copy=None)
generate the path in redirect_to instead of hardcoding it
Out of topic: add a functional test for it if there is none :)

BTW, I'm not sure why you use a Get parameter and not just a different URL for it??

score -1 · Accepted Answer · answered Oct 16 '13 at 21:14

-1

I ended up doing a redirect instead. That removes the get parameter.

answered Oct 16 '13 at 21:14

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How to remove get parameter from url in django view

2 Answers2