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In my HTML form with radio buttons and an 'other' text field the values of the radio buttons are not understood

I'm trying to create a HTML form which inserts all entered values into a MySQL database through php. Specifically for this case, the user enters a product number and a supplier .

The supplier should be defined by radio buttons and an additional ' other ' text field where the user can specify another option... When the user chooses the ' other ' option, the value typed there is inserted in the MySQL DB correctly, however when choosing one of the radio buttons, not the supplier is inserted but the text 'on'... I don't understand why the script does this, since I don't have the value 'on' anywhere in the script.

I've searched google and stackoverflow, and tried many things, but I'm still failing to get it to work. I'm a php and html newb so please go easy on me.

Below is a trimmed down version of the code. (In real, the script also submits more fields from the form, the session username, and a timestamp.)

Does anyone see where the 'on' may come from and how to solve this? Thank you.

The php part of the code:

<?php
    if (isset($_POST['submit'])) {
    // These set your initial variables. q1_* is for the queried entries in the form
    $q2_productname = $_POST['q2_productname'];
    $q3_supplier = $_POST['q3_supplier'];
    if ($q3_supplier =='other'){
    $q3_supplier = $_POST['q3_supplier_other'];
    }               


    if (isset($q2_productname)) {
    $q2_productname = trim($q2_productname);
    $q2_productname = strip_tags($q2_productname);
    $q2_productname = stripslashes($q2_productname);
    $q2_productname = htmlspecialchars($q2_productname);
    }
    if (isset($q3_supplier)) {
    $q3_supplier = trim($q3_supplier);
    $q3_supplier = strip_tags($q3_supplier);
    $q3_supplier = stripslashes($q3_supplier);
    $q3_supplier = htmlspecialchars($q3_supplier);
    }


    $dbc = @mysql_connect (DBHOST, DBUSER, DBPASS) or die('Failure: ' . mysql_error() );
    mysql_select_db(DBNAME) or die ('Could not select database: ' . mysql_error() );

    $query = "INSERT INTO orders VALUES ('$q2_productname','$q3_supplier')";
    $q = mysql_query($query);

    if (!$q) {
    exit("<p>MySQL Insertion failure.</p>");
    } else {
    mysql_close();
}

Then the html/echo/script part of the code:

<?php 
    if (isset($_POST['submit'])) {
    echo "<p style='padding: .5em; border: 2px solid red;'>Thank you. Your product has been added to the order list!</p>";
    }
?>

<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

<fieldset>
<div>
<label for="q2">Full product name:</label><br />
<textarea id="q2" name="q2_productname" rows="1" cols="50"></textarea>
</div>
</fieldset><br /><br />

<fieldset>
<div>
<label for="q3">Supplier (e.g. Sigma-Aldrich, VWR, ...):</label><br />

<input type="radio" name="q3_supplier" id="q3" onchange="disableTxt()" checked="checked" />Sigma-Aldrich 
    <input type="radio" name="q3_supplier" id="q3" onchange="disableTxt()" />VWR 
    <input type="radio" name="q3_supplier" id="q3" onchange="enableTxt()" />Other:
<input type="text" name="q3_supplier_other" id="other" disabled="disabled" />

<script>
    function disableTxt() {
        document.getElementById("other").disabled = true;
    }
    function enableTxt() {
        document.getElementById("other").disabled = false;
    }
</script>

</div>
</fieldset><br /><br />
</form>

Set the value attribute of your radio inputs. This value will be posted to your server. The value will be 'on' unless you specify it.

<input type="radio" name="q3_supplier" id="q3" onchange="disableTxt()" checked="checked" value='Sigma-Aldrich'/>Sigma-Aldrich 
<input type="radio" name="q3_supplier" id="q3" onchange="disableTxt()" value='VWR' />VWR 
<input type="radio" name="q3_supplier" id="q3" onchange="enableTxt()" value='Other' />Other:

The default value of input[type=radio] is on .

I suggest you learn how to use Firebug or Chrome dev tools, in order to actually see what is really submitted and what comes back from the server.

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