$_POST-ing issue..

[woolyg]

Hi all, I'm new to these forums. Hopefully someone can help.

I've created a dropdown menu by calling an array:

    <select name="prop_area" size="1" class="boxes">
        <?PHP
	$places = mysql_query ("SELECT * FROM places order by name");
	$ordered = array();
	while ($placelist = mysql_fetch_row($places)) {
		$id = $placelist[0];
		$name = $placelist[1];
		$ordered[$id] = $name;
	}
	natsort($ordered);
	while (list($id,$use) = each($ordered)) {
		$placeid = current($ordered); 
		$option = "<option value=\"$id\"";
		if ($id == $area) {
			$option .= " selected";
		}
		$option .= ">";
		$option .= "$use</option>";
		print $option;
	}
?>
    </select>

Now, what I want to do is use $_POST to process the selected item from the menu and insert it into a table. How would I do this? If I use

$prop_area = $_POST['prop_area'];

And then

$query = "INSERT INTO table (area) ".
"VALUES ('$prop_area')";

mysql_query($query) or die('Error, query failed : ' . mysql_error()); 

...it doesn't work!!! Can anyone help me?
Cheers,
WoolyG

[maxpup979]

After a cursory scan of the code, I don't see anything obvious (but I could easily be missing something)--so have you tried trouble shooting the query? if you print the query is the $prop_area variable being properly parsed?

f not, then what about in $_POST itself? Does that have the value you desire when you call the processing page?