Django: How do I update a model after a view has been rendered?

Question

I have a view that displays a list of models. Some of the properties of some of the models need to be updated after the view has been rendered: that is, the user is expected to see the original, unchanged values on first visit and the updated values on successive visits (or upon page reload).

I thought I could achieve this with class-based generic views. The official documentation hints at "doing some extra work before or after calling the generic view" (emphasis mine), but all examples given affect the models before it is displayed.

I looked into signals, but to no avail.

Spawning an asynchronous task is an option, but, since all I need to do is update a field in a few models (likely one or none) and save them, it seems overkill for the task at hand.

One could use an ajax request to trigger the update, or a custom template tag to display the relevant fields and update them thereafter. I dislike both, as they move application logic into the view layer. The ajax technique also adds the overhead of a second request.

Yet, I seem to have no other choices, or do I? Is there a more practical technique for plugging into the generic view or the request and execute extra logic after the template has been rendered?

score 9 · Accepted Answer · edited May 01 '20 at 10:52

There's no need for anything clever here. Rendering a template doesn't need to be the end of the view - that only comes when you actually return that as a response. They don't have to be in the same step.

So :

def my_view(request, params):
    #...do something...
    response = render_to_response('my_template.html', {'foo': bar})
    # ...do something after rendering...
    return response

Now, if you need to do this in a lot of views, you might want to do it in a decorator:

def my_decorator(view):
    def my_func(request, params):
        response = view(request, params)
        #...do something after rendering...
        return response
    return my_func

now you can decorate your views with @my_decorator and the action will be performed after rendering.

... or, it just occurred to me, if you want to do it in every view, you could do it in a middleware (just define the process-response method).

Edit after comment Class-based generic views use the new TemplateResponse class. You can add tasks that happen after the template is actually rendered with a post-render callback.

While that should work for the case where you do the view work manually, it doesn't seem to work if you delegate visualization to a generic view (I only tried with a class-based generic view, but function-based generic views are deprecated anyway), because on returning from the view the template has not yet been parsed, so it will see the updated models. — tawmas, Jun 01 '11 at 00:27

score 0 · Answer 2 · answered May 31 '11 at 21:52

0

Catch the return value from the view or function, do some processing, then kick it up the stack.

def foo(*args):
  ret = bar(*args)
  do_something()
  return ret

answered May 31 '11 at 21:52

Ignacio Vazquez-Abrams

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I tried that, but it doesn't work. When the original view returns, the template has not yet been parsed, so the modifications show up in the first visualization. – tawmas Jun 01 '11 at 00:18

score 0 · Answer 3 · answered May 31 '11 at 21:54

0

Could you grab two copies of the object, modifying and saving one while passing the other (unmodified and outdated) one to the view?

I haven't used class-based generic views yet, so I can't give you a specific example.

answered May 31 '11 at 21:54

Tim Yates

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That should work, but it implies passing a (potentially very long) list to the template. I'd prefer a solution with less overhead. – tawmas Jun 01 '11 at 00:20

score 0 · Answer 4 · answered May 31 '11 at 23:21

0

Django also comes with a built-in signal request_finished, that is sent AFTER django finished processing the request.

answered May 31 '11 at 23:21

Bernhard Vallant

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score 0 · Answer 5 · answered Jun 01 '11 at 03:04

If your project already use celery, you could create a task to update the model and and call

your_task.delay()

You could use the countdown argument to specify an elapsed before launching the task.

If you are not using celery chances are that you will in the future :)

Django: How do I update a model after a view has been rendered?

5 Answers5