CODEWITHSUNDEEP
jqueryajaxdjangotwitter-bootstrap
Bryce
Bryce

Reputation: 8712

How to build django ajax modal popup forms (with server side forms)?

What are good modern patterns or django apps for handling server side forms that appear in a popup? By this I mean:

  1. User action triggers a modal popup window.
  2. The form is server generated, demand loaded.
  3. In case of submit error, the server generates a new form.
  4. In case of success, there's a hook so DOM elements can be updated (the main page does not reload).

Assume common libraries like jquery or bootstrap 3 are available. Bonus points for DRY solutions: most patterns I've seen are kinda hairball, and the django ajax helpers I've looked into don't address the modal popup well or at all.

Upvotes: 6

Views: 9842

Answers (3)

Antony Fuentes
Antony Fuentes

Reputation: 1103

This is how I did it:

$(document).on('submit', 'div.modal-body form', function(e) {

    var form_el = $(this);

    e.preventDefault();
    $.ajax({
        type: $(this).attr('method'),
        url: $(this).attr('action'),
        data: $(this).serialize(),
        success: function (xhr, ajaxOptions, thrownError) {
            if ( $(xhr).find('.errorlist').length > 0 ) {

                form_el.parents('.modal-body').html(xhr);

            } else {

                form_el.parents('.modal-body').html('<h4>Form was properly processed!</h4>');

            }
        },
        error: function (xhr, ajaxOptions, thrownError) {

            form_el.parents('.modal-body').html(xhr);
        }
    });
});

So basically what this does is:

  • Watches the entire document for a submit event coming from a form element that is inside div.modal-body (bootstrap modal). If that event fires, then it gets captured in here.
  • I obtain the action and method using $(this), so no need to hard code
  • On success, if the result contains the class errorlist that means an error occurred, so instead of redirecting I capture that and I update the form to show the form including its error messages.
  • Otherwise, I update the modal saying that it was able to submit properly.

Upvotes: 1

Bryce
Bryce

Reputation: 8712

Here a way that uses plain jquery with a bootstrap styling the popup: See http://jsfiddle.net/brycenesbitt/afyY2/ . The trigger in a real app would look like:

$('#edit_foo').on('click', function () {
    $.ajax({
        type: 'GET',
        url: '/ajax/fooForm/',
        },
        success: function (data, textStatus, jqXHR) {
            $('#foo_modal').find('.modal-body').html(data);
            $('#foo_modal').modal('show');
        },
    });
});

The ajax call is expected to return the necessary form as html (no json).

On submit you'll need a hack because the jquery ajax error function won't return data: If the return code is 204 "no content" clean up and exit. If the return code is 200 redisplay the form with server side error notes.

See also this nice, but code-heavy, live form validation: http://www.the-art-of-web.com/javascript/ajax-validate/ and this grid of django ajax tools https://www.djangopackages.com/grids/g/ajax/ , none of which I particujlarly recommend.

Upvotes: 2

Bryce
Bryce

Reputation: 8712

There's a grid of Django ajax tools at https://www.djangopackages.com/grids/g/ajax/ you could look into.

Upvotes: 0

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