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Home/ Questions/Q 931831
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T20:34:03+00:00 2026-05-15T20:34:03+00:00

I have a html like below wherein there are multiple forms. I generate this

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I have a html like below wherein there are multiple forms. I generate this using JSTL, so the number could vary depending on what is enrolled in my DB. Each form has its own submit button.

Basically, I wanted to use the Form Plugins ajax submit button, but I don’t know how to reference the form.

<form id="form-1" action="approve.htm">
    <textarea name="comment"> </textarea>
    <input type="submit" value="Add Comment" />
</form>
<form id="form-2" action="approve.htm">
    <textarea name="comment"> </textarea>
    <input type="submit" value="Add Comment" />
</form>
.
.
.
<form id="form-10" action="approve.htm">
    <textarea name="comment"> </textarea>
    <input type="submit" value="Add Comment" />
</form>

My problem is how do I know which form id is getting submitted by using jQuery.

I don’t know which form-name will I place in the call to the ajax form.

$('#form-name?').ajaxForm();
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  1. Editorial Team
    Editorial Team
    2026-05-15T20:34:03+00:00Added an answer on May 15, 2026 at 8:34 pm

    By default it’ll submit the current form, e.g. the one you clicked “Add Comment” in, so just select them by tag name:

    $('form').ajaxForm();
    

    Alternatively (especially if there are other forms on the page), give them a class, for example:

    <form id="form-1" class="commentForm" action="approve.htm">
    

    And select by class:

    $('form.commentForm').ajaxForm();
    

    It’s important to keep in mind that .ajaxForm() doesn’t send the <form> it prepares it. You’re just selecting some forms and preparing them…they all still submit individually. So all you need to do here is use a selector that selects only the forms you want to prepare to submit through AJAX.

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