Question

I’m using the code found here (SO.com) to use the same template to both add and edit a record, but when I add a new record and click Submit, I get a 404 on the URL http://192.168.1.3:5678/app/student/edit/None/, and I’m not exactly sure why.

Here is the relevant portion of my urls.py:

    url(r'^app/lesson/new/$', 'edit_lesson', {}, 'lesson_new'),
url(r'^app/lesson/edit/(?P<id>\d+)/$', 'edit_lesson', {}, 'lesson_edit'),

Here is the relevant portion of my views.py:

def edit_lesson(request, id=None, template_name='lesson_edit_template.html'):
if id:
    t = "Edit"
    lesson = get_object_or_404(Lesson, pk=id)
    stu = get_object_or_404(Student, pk=sid)
    if stu.teacher != request.user:
        raise HttpResponseForbidden()
else:
    t = "Add"
    lesson = Lesson()

    if request.POST:
        form = LessonForm(request.POST, instance=lesson)
        if form.is_valid():
            form.save()

            # If the save was successful, redirect to another page
            return view_lessons(request)

    else:
        form = LessonForm(instance=lesson)

    return render_to_response(template_name, {
        'form': form,
        't': t,
        'lesson': lesson,
            }, context_instance=RequestContext(request))

And finally, here is my template:

<h1>{{ t }} Lesson</h1>
<form action="/app/lesson/edit/{{ lesson.id }}/" method="post"> {% csrf_token %}
{{ form.as_p }}
<input type="submit" value="Submit" />
</form>

I’m certain that I’m missing something really easy, but I can’t seem to put my finger on it. I’m using Django 1.3.1 if that makes any difference.

Thanks,
MC

Answer 1

There’s no need to specify any URL in the form’s action attribute. Just do

<form action="" method="post">

and it will POST back to the URL that you originally used to access it, which is what you want.