Novice here who learned to develop a web app with python using Flask. Now I'm trying to learn django 1.9 by redoing the same app with django. Right now I am stuck at trying to get the current URL and pass it as an argument so that the user can come back once the action on the next page is completed.
In Flask, to return to a previous URL, I would use the 'next' parameter and the request.url to get the current url before changing page.
In the template you would find something like this:
<a href="{{ url_for('.add_punchcard', id=user.id, next=request.url) }}">Buy punchcard :</a>
and in the view:
redirect(request.args.get("next"))
I thought it would be about the same with django, but I cannot make it work. I did find some suggestions, but they are for older django version(older than 1.5) and do not work anymore(and they are pretty convulsed as solutions goes!)
Right now, in my view I am using
return redirect(next)
Note: The use of return redirect in django seems very recent itself if I judge by solutions on the web that always seem to use return HttpResponse(..., so I take it alot of changes happened lately in how to do things.
and in the template I have
<a href="{% url 'main:buy_punchcard' member.id next={{ request.path }} %}">Buy punchcard</p>
but this actually return an error
Could not parse the remainder: '{{' from '{{'
I did add the context_processors in settings.py
TEMPLATE_CONTEXT_PROCESSORS = (
'django.core.context_processors.request',
)
But this is only the last error in a very long streak of errors. Bottom line is, I can't make it work.
As such, anyone could point me in the right direction as to what is the way to do this in django 1.9? It look like a pretty basic function so I thought it would be easier somehow.