Php Dynamic Menubar code error

Question

I am having a problem with making the links in my bar work properly the database is setup like so my db http://bloodkittens.com/resources/upload/db.png

<?php 
error_reporting(E_ALL);
ini_set('display_errors', '1');

mysql_connect("localhost", "dbuser", "******");
mysql_select_db("guild");

// prepare special array with parent-child relations
$menuData = array(
'items' => array(),
'parents' => array()
);
$result = mysql_query("SELECT  id_menu id, parentID_menu parentId, label_menu name FROM main_menu` ORDER BY parentID_menu"); 

while ($menuItem = mysql_fetch_assoc($result))
{
$menuData['items'][$menuItem['id']] = $menuItem;
$menuData['parents'][$menuItem['parentId']][] = $menuItem['id'];
}

// menu builder function, parentId 0 is the root
function buildMenu($parentId, $menuData)
{
$html = '';
$parent='';

if (isset($menuData['parents'][$parentId]))
{
    $menuClass= ($parentId==0) ? ' class="navbar" id="navbar"' : '';
    $parent= ($parentId==0) ? 0 : 1;    
    $html = "<ul{$menuClass}>\n";

    foreach ($menuData['parents'][$parentId] as $itemId)
    {
        //subment
        $result=mysql_query("select * from main_menu where parentID_menu='$itemId'");
        if (mysql_num_rows($result)>(int)0 && $parentId!=0) {
            $subm  =' class="navbar"';
        }else{
           $subm  ='';
        }
        //end

        $menu = $parentId == 0 ? ' class="menulink"' : '';      //class of main menu
        $html .= '<li>' . "<a{$subm}{$menu} href=\"#\" >{$menuData['items'][$itemId]['name']}</a>";

        // find childitems recursively
        $html .= buildMenu($itemId, $menuData);

        $html .= '</li>';
    }
    $html .= '</ul>';
}

return $html;
}

// output the menu
echo buildMenu(0, $menuData);
?>

How would i make it so that the value link_menu would be the href in the code for each separate entry in the db? rather then \'#\' because the code works completely and i'm very happy how it looks after i apply my css to it but the links aren't working

Saadulla · Accepted Answer · 2011-10-26 17:04:41Z

1

For work the menu sorting must be add a new field in the table, and change this query from

$result = mysql_query("SELECT id_menu id, parentID_menu parentId, label_menu name,  link_menu link FROM main_menu ORDER BY parentID_menu");

to :

$result = mysql_query("SELECT id_menu id, parentID_menu parentId, label_menu name,  link_menu link  FROM main_menu  ORDER BY menu_sort");

answered Oct 26, 2011 at 17:04

Saadulla

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Comments

miki · Accepted Answer · 2011-09-05 16:45:49Z

0

you never seem to select it

$result = mysql_query("SELECT id_menu id, parentID_menu parentId, label_menu, link_menu, name FROM main_menu ORDER BY parentID_menu");

and then..

$html .= '<li>' . "<a{$subm}{$menu} href=\"{$menuData['items'][$itemId]['link_menu']}\" >{$menuData['items'][$itemId]['name']}</a>";

answered Sep 5, 2011 at 16:45

miki

6953 silver badges8 bronze badges

2 Comments

Reza Over a year ago

this doesnt work because link_menu is not appart of $menuData. $menuData is an array of parrents and items. and those arrays are filled by $menuItems

Reza Over a year ago

Ok neither of your input fixed my problem but i fixed my code now so it all works

CodeCaster · Accepted Answer · 2011-09-05 16:46:03Z

0

Instead of href=\"#\" >, use this:

href=\"{$menuData['items'][$itemId]['link_menu']}\" >

answered Sep 5, 2011 at 16:46

CodeCaster

153k24 gold badges237 silver badges287 bronze badges

1 Comment

Reza Over a year ago

this doesnt work because link_menu is not appart of $menuData. $menuData is an array of parrents and items. and those arrays are filled by $menuItems

Reza · Accepted Answer · 2011-09-06 01:44:47Z

the code that i used was in the end like this i think i got it to work by pure luck but regardless i win : )

<?php 
error_reporting(E_ALL);
ini_set('display_errors', '1');

mysql_connect("localhost", "BloodKittens", "Zangoshi1");
mysql_select_db("bloodkittens");

// prepare special array with parent-child relations
$menuData = array(
'items' => array(),
'parents' => array(),
'links' => array()
);
 $result = mysql_query("SELECT id_menu id, parentID_menu parentId, label_menu name,  link_menu link FROM main_menu ORDER BY parentID_menu");

while ($menuItem = mysql_fetch_assoc($result))
{
$menuData['items'][$menuItem['id']] = $menuItem;
$menuData['parents'][$menuItem['parentId']][] = $menuItem['id'];
$menuData['links'] = $menuItem['link'];
}

// menu builder function, parentId 0 is the root
function buildMenu($parentId, $menuData)
 {
$html = '';
$parent='';

if (isset($menuData['parents'][$parentId]))
{
    $menuClass= ($parentId==0) ? ' class="navbar" id="navbar"' : '';
    $parent= ($parentId==0) ? 0 : 1;    
    $html = "<ul{$menuClass}>\n";

    foreach ($menuData['parents'][$parentId] as $itemId)
    {
        //subment
        $result=mysql_query("select * from main_menu where parentID_menu='$itemId'");
        if (mysql_num_rows($result)>(int)0 && $parentId!=0) {
            $subm  =' class="navbar"';
        }else{
           $subm  ='';
        }
        //end

        $menu = $parentId == 0 ? ' class="menulink"' : '';      //class of main menu
        $html .= '<li>' . "<a{$subm}{$menu} href=\"{$menuData['items'][$itemId]['link']}\" >{$menuData['items'][$itemId]['name']}</a>";

        // find childitems recursively
        $html .= buildMenu($itemId, $menuData);

        $html .= '</li>';
    }
       $html .= '</ul>';
   }

   return $html;
 }

// output the menu
echo buildMenu(0, $menuData);

?>

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