The target for ajaxSubmit is being duplicated instead of being replaced

I encountered a problem with the code below:

$('#refresh').click(function () {

    alert($('.report-container').length);

    $('.report-container').each(function () {

        var accordian = this;
        var url = $(this).children(':first-child').val();

        $('form').ajaxSubmit({
            url: url,
            success: function (responseText, statusText, xhr, $form) {
                $(accordian).html(responseText);
            }
        });

    });
});

This code is designed to refresh each of the tabular reports on the page using the user-configured form.

My expectation was that it would replace the one and only '.report-container' element with the fragment downloaded from the server. However, every time this runs, the call to 

alert($('.report-container').length);
increments?

This issue has been causing various problems - I'm not sure what I am missing here.

In addition to this, I have attempted to use the target property on the ajaxForm plugin with the same outcome.

I can confirm that the server responds with only one '.report-container' in the fragment, so it should be a one-to-one replacement.

javascriptjqueryajaxajaxform

Answer №1

Assuming you only have a single container and it will be replaced:

$('#update').on('click', function() {
    var box = $('.data-box');
    var link = box.children(':first-child').val();
    $('form').ajaxSubmit({
        url: link,
        success: function(response, status, xhr, $form) {
            box.replaceWith(response);
        }
    });
});

If you want to handle multiple containers, reintegrate the 'each' method but store the selection in a variable named "panels":

$('#update').on('click', function() {
    $('.data-box').each(function() {
        var panels = $(this);
        var link = panels.children(':first-child').val();
        $('form').ajaxSubmit({
            url: link,
            success: function(response, status, xhr, $form) {
                panels.replaceWith(response);
            }
        });
    });
});

Answer №2

If you need to swap out an element, simply utilize the .replaceWith() method.

The function html() is used to completely substitute the contents within the element. Therefore, post-replacement, your HTML structure will appear as follows:

<div class="container">
    <div class="new-container">

        ....

    </div>
</div>

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