I'm trying to build an active page menu with PHP and MySQL and am having a difficult time
fixing the error. In the while statement I have an if statement that is giving me fits. Basically I think I'm saying that "thispage" is equal to the "title" based on pageID and as the menu is looped through if "thispage" is equal to "title" then echo id="active".
Thanks
<?php
mysql_select_db($database_db_connection, $db_connection);
$query_rsDaTa = "SELECT * FROM pages WHERE pagesID = 4";
$rsDaTa = mysql_query($query_rsDaTa, $db_connection) or die(mysql_error());
$row_rsDaTa = mysql_fetch_assoc($rsDaTa);
$totalRows_rsDaTa = mysql_num_rows($rsDaTa);
$query_rsMenu = "SELECT * FROM menu WHERE online = 1 ORDER BY menuPos ASC";
$rsMenu = mysql_query($query_rsMenu, $db_connection) or die(mysql_error());
$thisPage = ($row_rsDaTa['title']);
?>
<link href="../css/MainStyle.css" rel="stylesheet" type="text/css" />
<h2><?php echo $thisPage; ?></h2>
<div id="footcontainer">
<ul id="footlist">
<?php
while($row_rsMenu = mysql_fetch_assoc($rsMenu)) {
echo (" <li" . <?php if ($thisPage==$row_rsDaTa['title']) echo id="active"; ?> . "<a href=\"../" . $row_rsMenu['menuURL'] . "\">" . $row_rsMenu['menuName'] . "</a></li>\n");
}
echo "</ul>\n";
?>
</div>
<?php
mysql_free_result($rsMenu);
mysql_free_result($rsDaTa);
?>
