如何發送ID到控制器笨

Question

我使用下面的代碼爲我的網站，但它顯示我

404：找不到網頁。

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來從URL中提取ID： 以下是在文檔的鏈接：http://www.codeigniter.com/userguide2/libraries/uri.html

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

Answer 2

我有可能創建一個簡單的它可以幫助你

這是我的控制器welcome.php從codeigniter新鮮

<?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'; 
    } 
} 

，這裏是我的路線

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

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

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

例如訪問它，如果工作

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

這個根據你的服務器是動態執行

見Pretty Url Setup CodeIgniter