I am trying to use the command foreach to create a drop down menu. That I can usually do, the problem is I am also adding some linking. Here is some my code:
<?sql

$sql2 = "SELECT mn.map, mn.route FROM maps m JOIN map_name mn ON (m.map_name = mn.map) WHERE `ship`='".$POST['ship']."'";
$result2 = mysql_query($sql2)
    or die("Query Error".mysql_error());
    while ($row = mysql_fetch_array($result2)) {
    $mapname[$row['mn.map']] = $row['mn.route'];
}
    ?>

<form action="edit_map_which.php" method="post">
<table border=0 cellspacing=1>
<tr>
<td>
Map :
</td>
<td>
<SELECT name="map">
<option value="" SELECTED>Select a Map</option>
<?php
foreach( $mapname as $map => $route ){ **
?>
<option value="<?php echo $map ?>"><?php echo $route ?></option>
<?php }
?>
</SELECT>
</td>
</tr>
<tr>
<td>
<INPUT type="submit" name="Submit" value="Submit">
</td>
</tr>
</table>

** I get the error Invalid argument supplied for foreach() for this line.

I've tried quite a few different things for this, and I now feel like I am beating my head against the wall. Thanks for the help.

"Invalid argument supplied for foreach()" indicates that your '$mapname' is not an array, which in turn implies that your SELECT is returning with an empty set. Have you tried echoing your SELECT statement and running it against your database to see what happens?

Take it easy,
Dan
Will-code-PHP-for-Athlons-and-powerbooks
"If Darl McBride was in charge, he'd probably make marriage unconstitutional too,
since clearly it de-emphasizes the commercial nature of normal human interaction, and
is a major impediment to the commercial growth of prostitution."