Question

我在我的网站上使用以下代码，但它显示了我

404：找不到页面。

路由

$route['class_name/function_name/(:num)'] = 'class_name/function_name/$1';

控制器

 public function function_name($Id)
    {
        print_r($Id); exit;
    }

Answer 1

使用URI从URL中提取id：以下是文档中的链接：http://www.codeigniter.com/userguide2/libraries/uri.html

示例 - How to get the a URL using uri_string() codeigniter?

Answer 2

我创造了一个简单的可能它可以帮助你

这是我的控制器welcome.php来自codeigniter

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

class Welcome extends CI_Controller {

    /**
     * Index Page for this controller.
     *
     * Maps to the following URL
     *      http://example.com/index.php/welcome
     *  - or -
     *      http://example.com/index.php/welcome/index
     *  - or -
     * Since this controller is set as the default controller in
     * config/routes.php, it's displayed at http://example.com/
     *
     * So any other public methods not prefixed with an underscore will
     * map to /index.php/welcome/<method_name>
     * @see http://codeigniter.com/user_guide/general/urls.html
     */

    public function index()
    {
        $this->load->view('welcome_message');
    }

    public function test1($ID) {
        echo $ID;
    }

    public function test2($ID, $ID2) {
        var_dump($ID, $ID2);
    }

    public function test() {
        echo 'test';
    }
}

这是我的路线

$route['default_controller'] = 'welcome';

$route['welcome/test1/(:num)'] = 'welcome/test1/$1';

$route['welcome/test/(:num)/(:num)'] = 'welcome/test/$1/$2';

工作时访问它的示例

https://192.168.248.209/stackoverflow/welcome/test2/1/3 https://192.168.248.209/stackoverflow/welcome/test1/1

根据您的服务器实现，这个是动态的

请参阅Pretty Url Setup CodeIgniter

如何在codeigniter

2 个答案: