How to render a Django template with a transient variable?

Question

I have an application that enforces a strict page sequence. If the user clicks the Back button, the application detects an out-of-order page access and sends the user back to the start.

I'd like to make this a bit more friendly, by redirecting the user back to the correct page and displaying a pop-up javascript alert box telling them not to use the Back button.

I'm already using a function that does a lot of validity checking which returns None if the request is okay, or an HttpResponseRedirect to another page (generally the error page or login page) if the request is invalid. All of my views have this code at the top:

response = validate(request)
if response:
    return response

So, since I have this validate() function already, it seems like a good place to add this extra code for detecting out-of-order access.

However, since the out-of-order detection flag has to survive across a redirect, I can't just set a view variable; I have to set the flag in the session data. But of course I don't want the flag to be set in the session data permanently; I want to remove the flag from the session data after processing the template.

I could add code like this to all of my render calls:

back_button = request.session.get('back_button', False)
response = render(request, 'foo.html', { 'back_button': back_button } )
if back_button:
    del request.session['back_button']
return response

But this seems a bit messy. Is there some way to automatically remove the session key after processing the template? Perhaps a piece of middleware?

I'm using function-based views, not class-based, btw.

Daniel Roseman Daniel Roseman · Accepted Answer · 2015-02-09T22:02:54

The session object uses the dictionary interface, so you can use pop instead of get to retrieve and delete the key at the same time:

back_button = request.session.pop('back_button', False)

How to render a Django template with a transient variable?

1 Answers