Django: render_to_response to another view

Question

How to pass the "render_to_response arguments" to an new view?

def show_list(request, id):
    try:
        profile = request.user.get_profile()
        list = Lista.objects.get(user_profile = profile)
    except Lista.DoesNotExist:
        c = {}
        c.update(csrf(request))
        titlepage = "ooops, you don't have a list YET! click on button to create one!"
        c = {'profile' : profile, 'request' : request, 'titlepage' : titlepage}
        return render_to_response('/profiles/list/create/',c,context_instance=RequestContext(request)) 

with this last line this doesn't work, the url /profiles/list/create/ redirects to a view create_list.

Well, I know that I could write something like redirect(/profiles/list/create/) but with this I cannot pass the dictionary c.

Thanks in advance

Answer 1

return render_to_response('my_template.html',
                      my_data_dictionary,
                      context_instance=RequestContext(request))

Answer 2

You should use django.contrib.messages to display your message and redirect user to another view.

View code:

from django.contrib import messages
from django.shortcuts import redirect

def show_list(request, id):
    try:
        # normal flow
    except Lista.DoesNotExist:
        messages.warning("ooops, you don't have a list YET! click on button to create one!")
        return redirect("/profiles/list/create/")

Somewhere in create template:

{% if messages %}
<ul class="messages">
    {% for message in messages %}
    <li{% if message.tags %} class="{{ message.tags }}"{% endif %}>{{ message }}</li>
    {% endfor %}
</ul>
{% endif %}

Also remember to properly enable messages.