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Home/ Questions/Q 416203
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T18:27:26+00:00 2026-05-12T18:27:26+00:00

I want to start processing some files from a django view and I want

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I want to start processing some files from a django view and I want to be able to send the name of the files to the browser as they are processed. Is there a way to do this (easily)? I could probably do this using threads and ajax calls, but I want the simplest solution for now.

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  1. Editorial Team
    Editorial Team
    2026-05-12T18:27:26+00:00Added an answer on May 12, 2026 at 6:27 pm

    I found what I needed in an answer from one of the links that Andre Miller provided.

    I found out that’s possible to pass an iterator to HttpResponse so I used this code and it worked:

    def import_iter():
        """ Used to return output as it is generated """
        # First return the template
        t = loader.get_template('main/qimport.htm')
        c = Context()
        yield t.render(c)
        # Now process the files
        if req.method == 'POST':
            location = req.POST['location']
            if location:
                for finfo in import_location(location):
                    yield finfo+"<br/>"
    
    return HttpResponse(import_iter())
    
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