Question

我正在尝试使用PHP Codeigniter创建动态菜单。

有一些主菜单每个主菜单都有一些Sub_Menu Like，Categories - ＆gt; Hardware，Software等。

我以某种方式制作但有问题，主菜单显示正确，但主菜单下的子菜单在每个主菜单中重复。

这是问题的形象：

我的观看代码

<ul class="slimmenu">
    <?php foreach($menus as $menu): ?>

    <li>
        <a href="#"><?php echo $menu -> Menu_Name;?></a>
        <ul>
        <?php foreach($submenus as $submenu): ?>
            <li>
                <a href="#"><?php echo $submenu -> Sub_Menu_Name;?></a>


            </li>

        <?php endforeach;?>
        </ul>
    </li>
    <?php endforeach;?>

</ul>

我的控制器代码

    <?php
defined('BASEPATH') OR exit('No direct script access allowed');

class MainSystem extends CI_Controller
{
    function __construct() {
        parent::__construct();
        $this->load->model('menusys','ms');
    }

    function menu()
    {
        $this->data['menus'] = $this->ms->get_menu();
        $this->data['submenus'] = $this->ms->get_submenu();
        $this->load->view('index', $this->data);
    }

    function add_menu()
    {
        $data = array(
            'Menu_Name' => $this->input->post('mname'),
            'Menu_Link' => $this->input->post('mlink'),

        );
        $this->ms->adm($data);
    }

    function sub_menu()
    {
        $this->data['mmenu'] = $this->ms->get_menu();
        $this->load->view('submenu', $this->data);
    }


    function add_submenu()
    {
        $data = array(
            'Main_Menu_Name' => $this->input->post('mmname'),
            'Sub_Menu_Name' => $this->input->post('mname'),
            'Sub_Menu_Link' => $this->input->post('mlink'),

        );
        $this->ms->adsm($data);
    }

}

?>

我的型号代码

<?php
defined('BASEPATH') or exit('no direct script directy allowed');

/**
 *
 */
class menusys extends CI_Model
{

    public function __construct()
    {
        parent::__construct();
    }

    public function adm($data)
    {
        $this->db->set('Is_Active',0);
        $this->db->insert('main_menu',$data);
        $this->session->set_flashdata('msg', 'Meun has been added');
        redirect('MainSystem/menu');
    }


    //sub-menu Section

    public  function get_menu()
    {

        //$this->db->where('Is_Active',1);
//        $this->db->select('*');
//        $this->db->from('main_menu');
//        $this->db->join('sub_menu', 'main_menu.Menu_Name = sub_menu.Main_Menu_Name','inner');
//        $query = $this->db->get();
//        return $query->result();
//
        $query=$this->db->get_where('main_menu',array('Is_Active'=>1));
        return $query->result();

    }
    function get_submenu()
    {
        $this->db->where('main_menu.Menu_Name = sub_menu.Main_Menu_Name');
        $this->db->select('*');
        $this->db->from('sub_menu');
        $this->db->join('main_menu', 'main_menu.Menu_Name = sub_menu.Main_Menu_Name','inner');
        $query = $this->db->get();
        return $query->result();


    }

    public function adsm($data)
    {
        $this->db->set('Is_Active',1);
        $this->db->insert('sub_menu',$data);
        $this->session->set_flashdata('msg', 'Meun has been added');
        redirect('MainSystem/menu');
    }


}

我想要的是什么：我希望每个Sub_Menu都会显示在他的父菜单下。

任何帮助将不胜感激。

更新

这是子菜单表图像。

Answer 1

您需要格式化菜单数组并将其循环播放

$menu = array(
    'menu_name_1' => array(
        'submenu1_1' => 'www.something.com',
        'submenu1_2' => 'www.something.com',
    ),
    'menu_name_2' => array(
        'submenu2_1' => 'www.something.com',
        'submenu2_2' => 'www.something.com',
    )
);

foreach($menu as $menu_name => $submenu){
    echo $menu_name.'<br>';
    if (!empty($submenu)){
        foreach($submenu as $submenu_name => $submenu_link){
            echo '<a href="'.$submenu_link.'">'.$submenu_name.'</a><br>';
        }
    }
}

您有包含菜单名称和子菜单的数组。首先你循环菜单名称和检查剂量它有子菜单，如果有，你循环子菜单并建立链接。

PHP Codeigniter

1 个答案: