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Home/ Questions/Q 591429
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T15:38:01+00:00 2026-05-13T15:38:01+00:00

I have the following code that works: <script type=text/javascript> $(document).ready(function() { // Initialise the

  • 0

I have the following code that works:

<script type="text/javascript">
$(document).ready(function() {
    // Initialise the table
    $('#table_1').tableDnD({
    onDrop: function(table, row) {
    $.tableDnD.serialize();

    $.ajax({
     type: "POST",
     url: "test.php?"+$.tableDnD.serialize(),
     data: "",
     success: function(html){
       alert("Success");
     }
    });
    }
});
});
</script>

Sending data to test.php:

<?php
$table_1[] = $_GET['table_1'];
$i = 0;
if(!empty($table_1[0])){
    foreach($table_1 as $value) {
        foreach($value as $row){
            $i++;
            mysql_query("UPDATE mytable SET tableOrder='$i' WHERE id = '$row'");
        }
    }
}
?>

As you can see the table_1 array retrieves the data using $_GET, but that ajax code says we’re sending with POST. If I change $_GET to $_POST it no longer works. Why is this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-13T15:38:02+00:00Added an answer on May 13, 2026 at 3:38 pm

    When you read from $_POST, you should pass the values in data instead of in the URL querystring.

    Your JavaScript code would have to change as follows:

    $.ajax({
     type: "POST",
     url: "test.php",
     data: $.tableDnD.serialize(),
     success: function(html){
       alert("Success");
     }
    });
    

    Then you would be able to do:

    <?php
    $table_1[] = $_POST['table_1'];
    ?>
    

    Your orignal code was working because as Mike Sherov noted in a comment below, any data passed in the URL querystring can always be accessed with $_GET, regardless of the HTTP verb used to submit the data.

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