Question

考虑这种情况:: - 当我在管理员中上传视图文件时，在用户端创建一个新菜单，即动态创建菜单..

例如：当管理员上传帮助视图文件时，帮助菜单应出现在用户端..就是这样可能，应该动态创建帮助控制器。

Answer 1

当然，PyroCMS会这样做。看一下模块管理代码，特别是details.php。你需要自己做一些开发，但我的代码将演示我是如何完成的。

http://github.com/pyrocms/pyrocms

Answer 2

我有这样的东西很容易回来......

看看我如何做到这一点的简单示例......这是一个帮助网站。

模特


<?php

class Template_model extends Model {

    function Template_model()
    {
        parent::Model();
    }

    function load_pages()
    {
        $data = array();
        $this->db->where('status',0);
        $this->db->order_by('sort', 'ASC');
        //$this->db->get('pages');
        $query = $this->db->get('pages');
        if ($query->num_rows() > 0){
            foreach ($query->result_array() as $row){
                $data[] = array(
                    "name" => $row['name'],
                    "url" => $row['url']
                );
            }
         }

        $query->free_result();
        return $data;
    }

}

控制器


<?php

class View extends Controller {

    function View()
    {
        parent::Controller();   
    }

    function index()
    {
        redirect('view/topic/orders');
        //echo "oops";
    }

    function topic()
    {
        $page = $this->uri->segment(3);

        //get the page id
        $this->load->model('view_model');
        $id = $this->view_model->get_page_id_by_url($page);
        $id = $id['id'];

        //get the page title
        $data['title'] = $this->view_model->get_page_title($id);

        //load the post by page id
        $data['posts'] = $this->view_model->get_page_posts($id);

        //load the header 
        $this->load->model('template_model');
        $data['pages'] = $this->template_model->load_pages();

        //load the view
        $data['main_content'] = 'view/page';
        $this->load->view('includes/template', $data);

    }

    function article()
    {
        $url = $this->uri->segment(3);

        //get the post by url
        $this->load->model('view_model');
        $data['post'] = $this->view_model->get_post_by_url($url);
        $id = $data['post']['page_id'];
        //print_r($data['post']);

        //get the page title
        $data['title'] = $this->view_model->get_page_title($id);

        //load the post by page id
        $data['posts'] = $this->view_model->get_page_posts($id);

        //load the header 
        $this->load->model('template_model');
        $data['pages'] = $this->template_model->load_pages();     

        //load the view
        $data['main_content'] = 'view/article';
        $this->load->view('includes/template', $data);
    }

}

/* End of file view.php */
/* Location: ./system/application/controllers/view.php */

和视图..


    <div id="navigation">
        <ul>

                <?php 
                if(!empty($pages))
                {   
                    foreach($pages as $page)
                    {
                        ?>

                            <li <?php if($on_page == $page['url']) { echo 'class="selected"'; }?>><a href="<?php echo base_url().'view/topic/'.$page['url'];?>"><?php echo $page['name'];?></a></li>

                        <?   

                    }
                }?>

                <!-- top navigation use selected class for selected item -->

                <div id="rNav" ><ul><li <?php if($on_page == 'admin') { echo 'class="selected"'; }?>><a href="<?php echo base_url();?>admin">Admin</a></li></ul></div>
                <div id="rNav" ><ul><li <?php if($on_page == 'logout') { echo 'class="selected"'; }?>><a href="<?php echo base_url();?>logout">Logout</a></li></ul></div>

        </ul>
    </div>

Answer 3

您应该将链接路由到处理菜单的特定控制器

在codeigniter中创建动态控制器？

3 个答案: