Code Examples -> PHP Nested Menu Example - Shows how to make a single level nested menu

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PHP Code Examples

Title :	PHP Nested Menu Example - Shows how to make a single level nested menu
Categories :	PHP, Navigation	John Harrison
Date :	Jul 27th 2002
Grade :	3 of 5 (graded 18 times)
Viewed :	30455
File :	No file for this code example.
Images :	No Images for this code example.

Search :	More code by John Harrison
Action :	Grade This Code Example
Tools :	My Examples List

Code

Used to create a single level nested menu.
simply call it like this: menu(get_page());
use this sql query to set up the table:

CREATE TABLE menu_data (
id int(11) unsigned NOT NULL auto_increment,
pid int(11) unsigned NOT NULL default '0',
title varchar(100) NOT NULL default '',
page_name varchar(100) NOT NULL default '',
PRIMARY KEY (id),
KEY pid (pid),
KEY page_name (page_name)
) TYPE=MyISAM;

I posted this because there arn't many nest menu examples availible on the net for php.

You will need to open a DB connection before using this. It should be easy to adapt
to other DBs like postgreSQL or MS-SQL

<?php
function get_page(){
$fn = explode("/", $_SERVER['PHP_SELF']);
$num_of_s = count($fn) - 1;
$fn = "$fn[$num_of_s]";
$query = "SELECT id, pid, title, page_name FROM menu_data WHERE page_name = '$fn'";
$result = mysql_query($query) or die("Query failed: $query " . mysql_error());
$num_results = mysql_num_rows($result);
$row = mysql_fetch_array($result);

return array($row['id'], $row['pid'], $row['title']);
}

function print_child($id, $sid){
$peer_query = "select * from menu_data where pid = $sid";
$peers = mysql_query($peer_query) or die("Query failed: $query " . mysql_error());
while ($prow = mysql_fetch_array($peers)){
if($id == $prow['id']){
echo '<img alt="CO2Busters menu tree" src="images/tree.gif" height="15" width="20">'.$prow['title'].' ';
}else{
echo '<img alt="CO2Busters menu tree" src="images/tree.gif" height="15" width="20"><a href="'.$prow['page_name'].'">'.$prow['title'].'</a> ';
}
}
}

function menu($menu_info) {
$query = "select * from menu_data order by id";
$result = mysql_query($query) or die("Query failed: $query " . mysql_error());;
$num_results = mysql_num_rows($result);

for ($i=0; $i < $num_results; $i++){
$row = mysql_fetch_array($result);
if($menu_info[0] == $row['id'] && $row['pid'] == 0){
echo $row['title'].' ';
print_child($menu_info[0], $menu_info[0]);
}elseif($menu_info[0] == $row['id'] && $row['pid'] == $menu_info[1]){
print_child($menu_info[0], $menu_info[1]);
}elseif($row['pid'] == '0') {
echo '<a href="'.$row['page_name'].'">'.$row['title'].'</a> ';
}
}
}
?>

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