Django - Link on submit - django

Question

I have a form with a POST Submit button. I want to link to a new page when the form is submitted and when the form.is_valid().
Is it better to make the link in the view or in the template? How to do it?
view.py:
from django.shortcuts import render_to_response
from ezmapping.models import *
from django.forms.models import modelformset_factory
def setName(request):
ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name'))
formset = ezAppFormSet(queryset=ezApp.objects.none())
if request.method == 'POST':
formset = ezAppFormSet(request.POST, request.FILES)
if formset.is_valid():
formset.save()
return render_to_response("project/manage_new.html", {'formset': formset, 'title': "New"}, context_instance=RequestContext(request))
template.html
{% extends "basemap.html" %}
{% block content %}
<table border="1">
<tr>
<td>
<h1>Define new App options</h1>
{% if formset.errors %}
<p style="color: red;">
Please correct the error{{ formset.errors|pluralize }} below.
</p>
{% endif %}
<form method="post" action="." encrypt="multipart/form-data">{% csrf_token %}
{{ formset.as_p }}
<input type="submit" value="Submit">
</form>
</td>
</tr>
</table>
{% endblock %}

Answer 1

You can use HttpResponseRedirect() in the view like this:
from django.shortcuts import render_to_response
from ezmapping.models import *
from django.forms.models import modelformset_factory
def setName(request):
ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name'))
formset = ezAppFormSet(queryset=ezApp.objects.none())
if request.method == 'POST':
formset = ezAppFormSet(request.POST, request.FILES)
if formset.is_valid():
formset.save()
next_link = u"/next/"
return HttpResponseRedirect(next_link)
return render_to_response("project/manage_new.html", {'formset': formset, 'title': "New"}, context_instance=RequestContext(request))

Answer 2

Sure. Just return a render_to_response inside of your conditional for request.method == 'POST', or optionally, just set up the context and template as variables to hand into render_to_response as such:
Paraphrased:
def foo(request):
template = 'template1.html'
# form and formsets here set into `context` as a dictionary
if request.method == 'POST':
template = 'template2.html'
return render_to_response(template, context)
[EDIT]
When I re-read your question, if you want to redirect to a different view if your form is valid, then return an HttpResponseRedirect instead.

Answer 3

If I understand your question correctly.
Your view:
from django.shortcuts import render_to_response
from ezmapping.models import *
from django.forms.models import modelformset_factory
def setName(request):
ezAppFormSet = modelformset_factory(ezApp, extra=1, fields=('name'))
formset = ezAppFormSet(queryset=ezApp.objects.none())
if request.method == 'POST':
formset = ezAppFormSet(request.POST, request.FILES)
if formset.is_valid():
formset.save()
submit_link = True # 2*
else:
submit_link = False
return render_to_response("project/manage_new.html", {'submit_link': submit_link, 'formset': formset, 'title': "New"}, context_instance=RequestContext(request))
Your template:
{% extends "basemap.html" %}
{% block content %}
<table border="1">
<tr>
<td>
<h1>Define new App options</h1>
{% if formset.errors %}
<p style="color: red;">
Please correct the error{{ formset.errors|pluralize }} below.
</p>
{% endif %}
<form method="post" action="." encrypt="multipart/form-data">{% csrf_token %}
{{ formset.as_p }}
<input type="submit" value="Submit">
</form>
{% if submit_link %}
<a href='/next/'>Data is saved, let's continue.</a>
{% endif %}
</td>
</tr>
</table>
{% endblock %}
Update.
And yes, if you want to redirect (not to show the next link) just place:
from django.shortcuts import redirect
...
return redirect( 'you-url' )
Instead of 2*.