注意：未定义的索引：index.php

Question

I' am getting `

Notice: Undefined index: index.php in xxxxx/index.php on line 26

When I remove the $submenu variable array from menu.php. Everything works fine but when I add $submenu array variable I get error message. ` What I' am trying to do here is get the top level first and then search if theirs any child submneu

On index.php file I have this code

    foreach($menu as $menu){
            $title = $menu[0];
            $privllages = $menu[1];
            $template = $menu[2];
            $class = $menu[3];
            $id = $menu[4];
            $icon = $menu[5];

            if(!$title){$title = 'No Title';}
            if($class){$class = 'class="'.$class.'"';} else {$class = '';}
            if($id){$id = 'id="'.$id.'"';} else {$id = '';}
            if(!$icon){$icon = "menu-default.png";}

            echo '<a href="'.$template.'" '.$class.' '.$id.'>' . $title . '</a>' . "<br/>";

/* LINE 26 Starts here */ 
                if($submenu[$template]){
                    foreach($submenu[$template] as $submenu){
                        echo '<a href="#">' . $submenu[0] . '</a>' . "<br/>";
                    }
                }
        }

On menu.php file

$menu[1] = array('Dashboard', '0', 'index.php', 'sidebar-menu','','menu-dashboard');
$menu[2] = array('Pages', '0', 'pages.php', 'sidebar-menu','','menu-dashboard');
$menu[3] = array('Media', '0', 'media.php', 'sidebar-menu','','menu-dashboard');
$menu[4] = array('Tools', '0', 'tools.php', 'sidebar-menu','','menu-dashboard');
$menu[5] = array('Users', '0', 'users.php', 'sidebar-menu','','menu-dashboard');
$menu[6] = array('Settings', '0', 'settings.php', 'sidebar-menu','','menu-dashboard');

$submenu['pages.php'][1] = array( 'All Links 1', 'admin-menu', 'manage_links', 'link-manager.php' );
$submenu['pages.php'][2] = array( 'All Links 2', 'admin-menu', 'manage_links', 'link-manager.php' );
$submenu['pages.php'][3] = array( 'All Links 3', 'admin-menu', 'manage_links', 'link-manager.php' );
$submenu['pages.php'][4] = array( 'All Links 4', 'admin-menu', 'manage_links', 'link-manager.php' );

Answer 1

Instead of

if($submenu[$template]){

use

if(!empty($submenu[$template])){

This will then only trigger if $submenu[$template] exists and is not empty