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Home/ Questions/Q 8203551
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Editorial Team
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Editorial Team
Asked: June 7, 20262026-06-07T07:35:47+00:00 2026-06-07T07:35:47+00:00

On the click of a button, I want to capture a form’s data and

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On the click of a button, I want to capture a form’s data and send it to a php page using ajax, json. Now, the alert “hello hello” comes on the click of a button ONLY WHEN I HAVE COMMENTED OUT MY AJAX PORTION.
When i include the ajax portion by un-commenting it, not only do i not get any errors but also my “hello hello” alert doesnt show up.
I find this odd. Why does it happen so?

Here is my ajax:

<script>
    $(document).ready(function () {
        $("#btn").click( function() {
        alert('hello hello');
        /* Coomenting starts here
        $.ajax({
            url: "connection.php",
            type: "POST",
            data: {
                id: $('#id').val(),
                name: $('#name').val(),
                Address: $('#Address').val()
            }
            datatype: "json",
            success: function (status)
            {
                if (status.success == false)
                {
                    alert("Failure!");
                }
                else 
                {
                    alert("Success!");
                }
            }
        });
        */
    //commenting ends here.
    //basically i just commented out the ajax portion 
    });
    });
</script> 
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-07T07:35:51+00:00Added an answer on June 7, 2026 at 7:35 am

    You are missing a comma after “data”:

    $.ajax({
        url: "connection.php",
        type: "POST",
        data: {
            id: $('#id').val(),
            name: $('#name').val(),
            Address: $('#Address').val()
        }, // Missing comma!
        datatype: "json",
        success: function (status)
        {
            if (status.success == false)
            {
                alert("Failure!");
            }
            else 
            {
                alert("Success!");
            }
        }
    });
    

    As @dystroy pointed out, since your code can’t compile, this malformed block will potentially prevent other code (such as a simple alert) from firing.

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