代码点火器自定义库

Question

我想建立一个自定义库，其中将包含将通过整个网站可用的功能。里面/application/libraries我做了一个新的文件CustomFuncts：

<?php if (! defined('BASEPATH')) exit('No direct script access allowed'); 
/** 
* @brief CommonFuncts  
* This class is used to provide common user functions to the whole application. 
* 
* @version 1.0 
* @date May 2012 
* 
* 
*/ 
class CommonFuncts extends CI_Controller { 

protected $ci; 

/** 
* function constructor 
*/ 
function __construct() 
{ 
    $this->ci =& get_instance(); 
} 
/** 
* @brief checkCookie 
* 
* Checks for a previous cookie, and if exists: 
* @li Loads user details to CI Session object. 
* @li Redirects to the corresponding page. 
* 
*/ 
public function verificaCookie(){ 
    $this->ci->load->view('index'); 
} 
} 
/* End of file CommonFuncts.php */ 
/* Location: ./application/libraries/CommonFuncts.php */ 

并在控制器welcome.php：

<?php if (! defined('BASEPATH')) 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->library('commonFuncts'); 
    $this->commonFuncts->verificaCookie(); 
} 
} 

/* End of file welcome.php */ 
/* Location: ./application/controllers/welcome.php */ 

我收到以下错误信息：

A PHP Error was encountered 

Severity: Notice 

Message: Undefined property: Welcome::$commonFuncts 

Filename: controllers/welcome.php 

Line Number: 23 

Fatal error: Call to a member function verificaCookie() on a non-object in 
/var/www/vhosts/unikodf.com/httpdocs/application/controllers/welcome.php on line 23 

我试着很多东西，包括在库中扩展CI_Controller和使用$this->load->view' instead of $ this-> ci-> load-> view`，但仍然收到相同的消息。 任何帮助表示赞赏。

Answer 1

如果库扩展CI_Controller。这意味着您实际上正在扩展控制器的功能。相反，声明你的图书馆这样的：

class CommonFuncts { } 

你并不需要从CI_Controller继承，因为你正在编写一个库不是一个控制器和一个图书馆并不是MVC模式的一部分。它是单个类，它扩展了框架的功能，修复了一个常见问题或者具有许多控制器使用的功能。

要访问它的方法使用：

$this->load->library('CommonFuncts'); 
$this->commonfuncts->verificaCookie(); 

如果你想简化与调用库使用的名称：

// The second argument is the optional configuration passed to the library 
// The third argument is the name with which you would like to access it 
$this->load->library('CommonFuncts', '', 'common'); 
$this->common->verificaCookie(); // /\ = configuration 
//  /\/\/\ = name 

Answer 2

你很困惑controllers与libraries。

自定义库不应该扩展CI_Controller。 The documentation outlines how to create your own custom library。

其次，你需要加载库，然后才能访问它的方法：

$this->load->library('CommonFuncts'); 
// The "CommonFuncts" library is now ready to use 
$this->commonFuncts->verificaCookie(); 

Answer 3

这里有几个不错的教程在CI

Creating Libraries

Using CodeIgniter Libraries

创建自定义库

CodeIgniter - Form: Creating Menu Library

Building with CodeIgniter: Using Libraries

Creating a simple, extensible CodeIgniter authentication library

Creating your own library in CodeIgniter

Answer 4

在这里，我要修改一件事（我正在使用代码点火器2.1。3）：

 
//echo CI_VERSION; to get the code igniter version 
Plesae use 
$this->commonfuncts->verificaCookie(); //f is small here commonfuncts 
instead of 
$this->commonFuncts->verificaCookie();

Answer 5

您的构造函数不正确。您使用：

$this->ci =& get_instance(); //This is invalid syntax. 

分配codeIgniter一个变量，并且使用该变量对CodeIgniter的资源

class exampleClass{ 
    protected $codeIgniter; 

    public function __construct() 
    { 
     $this->codeIgniter =& get_instance(); 
     $this->codeIgniter->load->helper('url'); 
     $this->codeIgniter->config->item('base_url'); 

    } 
}