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Asked: 2026-05-29T06:13:22+00:00

I have these codes: testing.php <html> <head> <script type=text/javascript> function copy_data(id){ var a =

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I have these codes:

testing.php

<html>
<head>
<script type="text/javascript">
        function copy_data(id){
            var a = document.getElementById(id).value;
            document.getElementById("copy_to").value=a;
        }
    </script>
</head>
<body>
<form action="testprocess.php" method="post">
<input type="text" name ="a" id="copy_from" onkeyup="copy_data('copy_from')"/>
<input type="text" value=0 name ="b" id="copy_to"/>
<input type="submit">
</form>
</body>
</html>

testprocess.php

<?php

    $test = $_POST['copy_to'];

    echo $test;
?>

I get an error saying that 'copy-to' is an undefined variable. Can you please tell me why?
Thanks.

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1 Answer

  1. Editorial Team

    $_POST values are passed through an element’s name attribute rather than the ID. Try this:

    <input type="text" value=0 name="copy_to" id="copy_to"/>

    And make sure you use a an underscore in your PHP variable:

    $test = $_POST['copy_to'];
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