Navigation bij menu with query

[Martijn_Apon]

Hello to all,

My first post and I hope I didn't miss a topic about the same problem(cause there so many topics)

I've got a little problem with a menu : navigationmenu

The menu is a combination of css en php and a mysql tabel crated with the following code:

function Navigatie()
{
	echo "<div class='menu'>";
	echo"<ul id='nav'>\n";
    $menu=mysql_query("select * FROM artikelgroep "); // Laat de Artikelgroepen zien die er zijn (hoofdgroepen muziekinstrumenten)
    while ($record=mysql_fetch_row($menu))
	 {
	   echo "<li><a class='hide' href='#'><b>$record[1]</b><!--[if IE 6]><!--></a><!--<![endif]-->\n";
	   echo "<ul>\n";
	   $submenu=mysql_query("select * FROM artikelsubgroep WHERE (Artikelgroep = '$record[0]')"); 
       while ($record=mysql_fetch_row($submenu))
			{
			echo"<li><a href='#'>".$record[2]." </a></li>\n"; // Dropdown from items  (subgroepsoorten)
			// *** I think here should be some kind of clickable and general query to create automatically a list of the selected item			}
	   echo "</ul>\n";	
	   echo "</li>\n";
     }
    echo"</ul>\n";
    echo "</div>";
}

What I want is:
Instead a link to a php page with a query to show a list of the selected item is a general query that shows the list of the selected items. So it prevents a growing number of php pages and something else if in a cms a new item is created i don't need toe create an onther query.

I hope I've ben clear enough to someone who can help me with this problem.

Thanks already

Greetz Martijn

[suomynonA]

while ($record=mysql_fetch_row($submenu))
            {
            echo"<li><a href='#'>".$record[2]." </a></li>\n"; // Dropdown from items  (subgroepsoorten)
            // *** I think here should be some kind of clickable and general query to create automatically a list of the selected item            } 

I think you might be looking for something like this:

echo("<select name='dropdown'>");
while ($record=mysql_fetch_row($submenu))
{
  echo("<option value='".$record[2]."'>".$record[2]."</option>");
}
echo("</select>");

However, this should be included in a form, which, once submitted, will pass the value of the selected item.

Of course you can write the pure html parts (such as the select tag) outside php code, I just wrote the first code that came across my mind.

[Martijn_Apon]

I managed it by doing the following:

function Navigatie()
{
	echo "<div class='menu'>";
	echo"<ul id='nav'>\n";
    $menu=mysql_query("select * FROM artikelgroep "); // Laat de Artikelgroepen zien die er zijn (hoofdgroepen muziekinstrumenten)
    while ($record=mysql_fetch_row($menu))
	 {
	   echo "<li><a class='hide' href='#'><b>$record[1]</b><!--[if IE 6]><!--></a><!--<![endif]-->\n";
	   echo "<ul>\n";
	   $submenu=mysql_query("select * FROM artikelsubgroep WHERE (Artikelgroep = '$record[0]')"); 
       while ($record=mysql_fetch_row($submenu))
			{
			echo"<li><a href='lijst.php?id=".$record[0]."'>".$record[2]." </a></li>\n"; // Dropdown van instrumenten (subgroepsoorten)
			// *** Volgens mij moet hier een query komen om automatisch een link te maken met betreffende onderwerp
			// *** Maar welke, ik kom er niet uit?
			}
	   echo "</ul>\n";	
	   echo "</li>\n";
     }
    echo"</ul>\n";
    echo "</div>";
}



Function ShowProducts()
{
	$Subgroepsoort = $_GET['id']; // Neemt Id mee om in de query de juiste uitlezing te kunnen doen

	$produkten = mysql_query("select * FROM artikel WHERE (Subgroepsoort = '$Subgroepsoort')");
	echo "<table border=1>";
	echo "<tr><td></td><td>Omschrijving</td><td><img src='plaatjes/van.gif'</td><td><img src='plaatjes/voor.gif'</td>";
	while ($record=mysql_fetch_row($produkten))
	{
	 echo"<tr><td><img src='plaatjes/".$record[7]."'</td><td><b>".$record[3]."</b><br>".$record[4]."</td><td>".$record[5]."</td><td> ".$record[6]."</td></tr>" ; 
	}
	echo "</table>";
}

Thanks
Martijn