$_POST Array not being passed

suttercain · December 5, 2007

Hi guys,

I have two select menus using javascript and the first one is populated via a mysql query. The idea is simple, select items from the left menu and add them to the right menu. I then want to inser the information from the right select menu into the database. The problem I am having is that the information that I am adding to the right select menu (or even the left menu for this matter) is not being sent to the $_POST variable.

<body>

<?php
if (isset($_POST['submit'])) {
echo $_POST['sel2'];
print_r($_POST);
}


$sql = mysql_query("SELECT * FROM characters ORDER BY name");

?>
<form action="<?php $_SERVER['PHP_SELF']; ?>" method="post">
<table border="0">
<tr>
	<td>
		<select name="sel1" size="10" multiple="multiple">
            <?php
		while ($row = mysql_fetch_array($sql)) {
		echo "<option value='".$row['character_id']."'>".$row['name']. " - ".$row['alter_ego']. "</option>";
		}
		?>
		</select>
	</td>
	<td align="center" valign="middle">
		<input type="button" value="-->"
		 onclick="moveOptions(this.form.sel1, this.form.sel2);" /><br />
		<input type="button" value="<--"
		 onclick="moveOptions(this.form.sel2, this.form.sel1);" />
	</td>
	<td>
		<select name="sel2" size="10" multiple="multiple">

		</select>
	</td>
</tr>
</table>
<input type="submit" name="submit">
</form>
<?php

?>

</body>

I can rationalize why the content from the second select menu, sel2, is not being passed, because there are no values. But shouldn't the first select menu be passed?

When I submit the form and print_r($_POST) I get:

Array ( [submit] => Submit Query )

Any help would be appreciated..

Thanks.

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$_POST Array not being passed

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