Django How to stop the execution of a function

Question

I am trying to call a python function by Ajax. I want to stop the execution of that function. I have tried to refresh the page but it doesn't stop. When I stopped the Django server then it get stop.

Here some code: This Call call HTTP request

class ScrapData():

    @classmethod
    def search_status(self, urls_list):
      for url in urls_list:
            r = requests.get(url)
            text = r.text
            #...

      return 1

Calling this view by Ajax

class SearchData(View):

    def get(self, request, *args, **kwargs):
        urls_list = [];
        response = ScrapData().search_status(urls_list)

        return HttpResponse(response)

I want to stop the function. How can I do this? Thanks

What event is supposed to stop your function? Probably, you could set timeout for your request? — Fomalhaut, Jan 09 '17 at 07:18

score 1 · Answer 1 · edited May 23 '17 at 11:52

1

How is this django app served? How many processes and threads? How do you know which thread runs the particular request? There are several Python ways to deal with this, but it is not as simple as "stop a particular request".

In my opinion, the best way to deal with long-polling tasks in terms of management, scaleability and self-sanity is to use some tool like Celery. There is also a dedicated SO question on how to revoke a Celery task.

Django Channels is another fresh approach to the matter, but it needs a bit of more development yet before it can be used rather than Celery.

edited May 23 '17 at 11:52

Community

1
1

answered Jan 09 '17 at 07:17

Wtower

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thanks, is there any way to handle it by session, or through database etc? – Shoaib Ijaz Jan 09 '17 at 07:50
Welcome. Sure, you deal with those within the view that is called when the cancel ajax is sent. – Wtower Jan 09 '17 at 08:48

score 0 · Answer 2 · answered Oct 26 '17 at 04:16

0

I got the same problem that I called a a wrong and slow request. And I stopped it by restarting my Apache server. It might be a solution when it's ok to restart the server.

answered Oct 26 '17 at 04:16

JohnTsai

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score 0 · Answer 3 · answered Mar 02 '21 at 14:41

try nested function, it will allow you to call whichever function you wish, or end process

the code:

class ScrapData():

    @classmethod
    def func():
        def search_status(self, urls_list):
              for url in urls_list:
                r = requests.get(url)
                text = r.text
                #...

          return 1
        def get(self, request, *args, **kwargs):
            urls_list = [];
            response = ScrapData().search_status(urls_list)

            return HttpResponse(response)
        def empty():
            pass
    #call the function you want to be excute further on
    #if you want nothing to happen then call empty function(pass keyword)

did this answer help? pls tell or at least accept – Mar 04 '21 at 08:59 — , Mar 04 '21 at 08:59

Django How to stop the execution of a function

3 Answers3