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Home/ Questions/Q 7031379
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Editorial Team
  • 0
Editorial Team
Asked: May 28, 20262026-05-28T00:47:44+00:00 2026-05-28T00:47:44+00:00

I have some jQuery that takes the value of a text input and puts

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I have some jQuery that takes the value of a text input and puts it into a MySQL database. However, when the jQuery runs, the page refreshes and the variables in the form appear in the URL almost as GET variables. However, none of the variables are GET. Ideally, I would like the page not to refresh.

jQuery:

$('.commentBox').keypress(function(e) {

    if(e.which == 13) {
        if ($.trim($(this).val()) == ""){
            $('#nocomment').modal('show');
        }
        else {
            var form = $(this).siblings('.commentForm'); 
            var commentbox = $(this).val();

            $.ajax({
                type: "POST",
                url: "../comment",
                data: form.serialize(),
                success: function(){
                    commentbox.val('');
                    form.siblings('.commentContainer').append(response);
                } 
            });
        }
    }

});

HTML (echoed from PHP):

<form class='commentForm'>
    <input type='hidden' name='record_id' value='$answerid[$f]' />
    <input type='hidden' name='question_id' value='$q' />";
    <input type='text' class='commentBox' placeholder='...comment' name='comment' autocomplete='off' />";
</form>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-28T00:47:44+00:00Added an answer on May 28, 2026 at 12:47 am

    You have to either return false or prevent default, which will stop the form from submitting:

    $('.commentBox').keypress(function(e)
    {
        if(e.which == 13)
        {
            e.preventDefault(); // <-- This will stop the form from submitting.
    
            if ($.trim($(this).val()) == "")
            {
                $('#nocomment').modal('show');
            }
            else
            {
                var form = $(this).closest('.commentForm');
                var commentbox = $(this).val();
                $.ajax({
                    type: "POST",
                    url: "../comment",
                    data: form.serialize(),
                    success: function(){
                        commentbox.val('');
                        form.siblings('.commentContainer').append(response);
                    }
                });
            }
        }
    });
    
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