active/current links for sub menu in PHP on CSS menu - php

Question

I have the below code for my CSS menu, it gets all the links from a database using PHP.
I have set active/current links for selected items which works fine for top level links but where i have sub menus it doesnt work.
for example, i have the following:
Services
- Service 1
-- Sub Service 1
If Service 1 is selected, the Services link is active/current but if the Sub Service 1 link is selected, i want the Services link to be active/current but its not
how can i fix this using the below code?
<ul class="nav">
<?php
//select all the top row items
$sql="SELECT * from website_menu where parent_top = '' and parent = '' order by sequence ASC ";
$rs=mysql_query($sql,$conn) or die(mysql_error());
while($result=mysql_fetch_array($rs))
{
//then select all the next rows down (parent_top)
$current = false;
$subMenu = '';
$sql2="SELECT * from website_menu where parent_top = '".$result["sequence"]."' order by sequence ASC ";
$rs2=mysql_query($sql2,$conn) or die(mysql_error());
if(mysql_num_rows($rs2) > 0)
{
$subMenu = '<ul>';
while($result2=mysql_fetch_array($rs2))
{
if($_GET["id"] == $result2["link"])
{
$current = true;
}
$subMenu .= '<li><span>'.$result2["title"].'</span>';
//
$sql3="SELECT * from website_menu where parent = '".$result2["sequence"]."' ";
$rs3=mysql_query($sql3,$conn) or die(mysql_error());
if(mysql_num_rows($rs3) > 0)
{
$subMenu .='<ul>';
while($result3=mysql_fetch_array($rs3))
{
$subMenu .='<li><span>'.$result3["title"].'</span></li>';
}
$subMenu .='</ul>';
$subMenu .='</li>';
}
else
{
$subMenu .='</li>';
}
}
$subMenu .= '</ul>';
}
echo '<li';
if($_GET["id"] == $result["link"] || $current)
{
echo ' class="active"';
}
echo '><span>'.$result["title"].'</span>', $subMenu, '</li>';
}
?>
</ul>

Answer 1

You might want to try this
In your following while loop make this change
while($result3=mysql_fetch_array($rs3))
{
$subMenu .='<li><span>'.$result3["title"].'</span></li>';
// add this if block
if($_GET["id"] == $result3["link"])
{
$current = true;
}
}