带有ID和标题的CodeIgniter URL路由

Question

我是CodeIgniter的新手，我想把我的网址从

http://test.com/1/test-post

至

http://test.com/test-post-1

在CI有路线。如何通过URL路由实现这一点？

我的表发布了

posts_id
posts_name

My Home View (调用这些帖子)

我使用url标题作为url：

<a href="<?php echo base_url("{$v['posts_id']}/".url_title($v['posts_name'], "-", true)) ?>">

我的职位控制器

    <?php 
defined('BASEPATH') OR exit ('No direct script access allowed');
class Posts extends CI_Controller {
    public function __construct()
    {
        parent::__construct();
        $this->load->model('Posts_model');
    }
    public function index()
    {
        show_404();
    }
    public function details($posts_id=NULL,$posts_name='') {
        $data['title'] = $this->Posts_model->select_content_by_id($posts_id);
        $data['data'] = $this->Posts_model->select_content_by_id($posts_id);
        $this->load->view('posts',$data);
    }
}

我的帖子--模型

 <?php 
defined('BASEPATH') OR exit ('No direct script access allowed');
class Posts_model extends CI_Model{
    public function dsPosts()
    {
        return $this->db->get('posts')->result_array();
    }
    public function select_content_by_id($posts_id)
    {
        return $this->db->where('posts_id',$posts_id)->get('posts')->result_array();
    }
}

我的路线

$route['default_controller'] = 'home';
$route['404_override'] = '';
$route['(:num)'] = 'posts/details/$1';
$route['(:num)/(:any)'] = 'posts/details/$1/$2';

Answer 1

你已经接近答案了

$route['(:any)-(:num)'] = 'posts/details/$2/$1';

然后在视野中

<a href="<?php echo base_url(url_title($v['posts_name'], "-", true)."-{$v['posts_id']}".) ?>">

就这样。但我建议您写得更清楚，因为您可以有多个控制器！如果您有注释控制器并希望从它获取数据，会发生什么？(:any)意味着它可以是任何东西。