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Home/ Questions/Q 3333800
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Editorial Team
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Editorial Team
Asked: May 17, 20262026-05-17T23:50:10+00:00 2026-05-17T23:50:10+00:00

I am trying to change my code to use json on recommendation from a

  • 0

I am trying to change my code to use json on recommendation from a previous question to simplify things a bit…

On client side:

<script type="text/javascript" src="../js/jquery-1.3.2.min.js"></script>
<script type="text/javascript" src="../js/jquery.tablednd_0_5.js"></script>
<script type="text/javascript" src="../js/jquery.json-2.2.js"></script>

<script type="text/javascript">
    $(document).ready(function() {
        $('#table').tableDnD();
    });
    function sendData() {
        data = $('#table').tableDnDSerialize();
        alert(data); // shows expected data
        document.dataform.data.value = $.toJson(data);
        document.data.submit();
    }
</script>

<form action="$php_page_name" method="post" name="dataform" onSubmit="sendData()">
    <input type="hidden" name="data" />
    <input type="submit" value="Submit" />
</form>

The js alert outputs the expected array, which i think is converted to a string by this point. But when I submit form.data, my php:

$data = json_decode($_POST['data']);
print_r($data);
print_r($_POST);

returns only:

Array ( [data] => )

Any ideas why nothing is being passed ?

Cheers,
Andy

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-17T23:50:10+00:00Added an answer on May 17, 2026 at 11:50 pm

    You need to wait for the return of the function:

    onSubmit="return sendData()"
    

    Otherwise the form will be submitted immediatly and does’nt wait till data is changed.

    inside the function replace this

    document.data.submit();
    

    with this:

    return true;
    

    Furthermore: assuming you use this as jquery.json-2.2.js :
    http://code.google.com/p/jquery-json/downloads/detail?name=jquery.json-2.2.js&can=2&q=
    The method-name is
    $.toJSON instead of $.toJson

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