Member Avatar for samergamer

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."'";
$result2 = mysql_query($sql2)
or die("Query Error".mysql_error());
while ($row = mysql_fetch_array($result2)) {
$mapname[$row] = $row;
}
?>

<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.

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  • 20 Hours Discussion Span
  • Latest Post Latest Post by Roberdin
Member Avatar for Roberdin

Sounds like that your query is pulling nothing. What I would do, is above the while() section of your code, add

$mapname = array();

This will force $mapname to be an array - otherwise, if your query pulls nothing, then the while() block won't be interpreted and hence your mapname is undefined. So when foreach() asks for it, it finds that its not an array and starts whining. Forcing $mapname to be an array will just mean that the foreach doesn't do anything if its empty.
Obviously, you need to put this code above wherever else that you may have used $mapname as it will destroy it first.
I do, however, have some concerns about your query.

`ship`='".$POST['ship']."'";

Should ship, not in fact be, m.ship or mn.ship?

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