Question

我正在接受`

注意：未定义的索引：index.php in 第26行的xxxxx / index.php

当我从menu.php中删除$ submenu变量数组时。一切正常，但当我添加$ submenu数组变量时，我收到错误消息。 ` 我在这里要做的是首先获得最高级别，然后搜索他们的子任务是否为子项目

在index.php文件中我有这段代码

    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/>";
                    }
                }
        }

在menu.php文件

上

$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

而不是

if($submenu[$template]){

使用

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

这将仅在$ submenu [$ template]存在且不为空时触发

Answer 2

尝试使用此修改后的代码。代码中提到了错误

<?php
$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' );

foreach($menu as $arr_menu){ // problem-1 was  here foreach($menu as $menu)
        $title = $arr_menu[0];
        $privllages = $arr_menu[1];
        $template = $arr_menu[2];
        $class = $arr_menu[3];
        $id = $arr_menu[4];
        $icon = $arr_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.'>' . $title . '</a>' . "<br/>";
            if(isset($submenu[$template])){  // problem-2 was  here not checking isset
                foreach($submenu[$template] as $arr_submenu){ // problem-3 was  here foreach($submenu as $submenu)
                    echo '<a href="#">' . $arr_submenu[0] . '</a>' . "<br/>";
                }
            }
    }
?>

Answer 3

更改

if($submenu[$template])

到

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

以下行也不明确

foreach($menu as $menu)

将其更改为

foreach($menu as $my_menu)

注意：未定义的索引：index.php

3 个答案: