CODEWITHSUNDEEP
htmldjangoforms
tvg
tvg

Reputation: 23

Submit form on Django page without changing url

I have a text area in a django based website that displays the content of a database field. I'd like to be able to edit this field and submit it to a function which updates the field in the database.

I know how I can call a function in a views.py which then send back the result of the query using render_to_response to a new webpage.

So in summary, how do I adress a function in a django/python script using a html form without the need to refer to another url?

Upvotes: 2

Views: 5031

Answers (3)

super9
super9

Reputation: 30111

This is the standard views code pattern that I've been using.

def payment_details(request, obj_id):
    yourobj = get_object_or_404(Obj, pk=obj_id)
    form = TheForm(instance=yourobj)

    if request.method == 'POST':
        form = TheForm(request.POST, instance=yourobj)
        if form.is_valid():
            yourobj = form.save()
            messages.success(request, 'Yourobj is saved!')
            url = reverse('SOMEURL')
            return redirect(url)

    template = 'SOMETEMPLATE'
    template_vars = {'TEMPLATEVARS': TEMPLATEVARS}
    return render(request, template, template_vars)

Having watched the Advanced Forms talk at DjangoCon, one could re-write the above view like this:

def payment_details(request, obj_id):
    yourobj = get_object_or_404(Obj, pk=obj_id)
    form = TheForm(request.POST or NONE, instance=yourobj)

    if request.method == 'POST' and form.is_valid():
        yourobj = form.save()
        messages.success(request, 'Yourobj is saved!')
        url = reverse('SOMEURL')
        return redirect(url)

    template = 'SOMETEMPLATE'
    template_vars = {'TEMPLATEVARS': TEMPLATEVARS}
    return render(request, template, template_vars)

Upvotes: 1

BluesRockAddict
BluesRockAddict

Reputation: 15683

It's usually recommended to use Post/Redirect/Get pattern, for example:

def myview(request, **kwargs):

    if request.POST:
        # validate the post data

        if valid:
            # save and redirect to another url
            return HttpResponseRedirect(...)
        else:
            # render the view with partial data/error message


    if request.GET:
        # render the view

        return render_to_response(...)

Upvotes: 4

Chris Pratt
Chris Pratt

Reputation: 239290

Use AJAX:

1) Create a view to handle form submission:

def my_ajax_form_submission_view(request):
    if request.method == 'POST':
        form = MyForm(request.POST)
        if form.is_valid():
           # save data or whatever other action you want to take
           resp = {'success': True}
        else:
           resp = {'success': False}

        return HttpResponse(simplejson.dumps(resp), mimetype='application/json')

    else:
        return HttpResponseBadRequest()

Then, tie the view into your urlpatterns

2) Submit form via AJAX (uses jQuery):

$('#my-form-id').submit(function(){
    var $form = $(this);
    $.post('/url/to/ajax/view/', $form.serialize(), function(data, jqXHR){
        if (data.success) {
            alert('Form submitted!');
        } else {
            alert('Form not valid');
        }
    });
    return false;
});

That's the basics. You can and should provide more detailed return responses, error handling, form validation/checking, etc.

Upvotes: 3

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