php - 将变量值从库传递到 Controller ，然后在 CI 3.0 中查看

Question

我正在尝试将变量值从库函数传递到 Controller ，然后查看以获取输出。

为此，我的库代码:

<?php                                                                                   

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

/**
 * @description : Library to access MyOperator Public API
 */
Class My_Operator extends Admin_controller{

    protected $developers_url = 'https://developers.myoperator.co/';
    protected $token = 'XXXXXXXXX';

    function __construct() {

    }

    public function run() {
        # request for Logs
        $url = $this->developers_url . 'search';
        $fields = array("token" => $this->token);
        $result = $this->_post_api($fields, $url);

        $this->log("result");
        $this->log($result);
    }

    private function _post_api(Array $fields, $url) {
        try {
            $ch = curl_init();
            curl_setopt($ch, CURLOPT_URL, $url);
            curl_setopt($ch, CURLOPT_TIMEOUT, 30);
            curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
            curl_setopt($ch, CURLOPT_CUSTOMREQUEST, 'POST');
            curl_setopt($ch, CURLOPT_POSTFIELDS, $fields);
            $result = curl_exec($ch);
        } catch (Exception $e) {
            return false;
        }
        $this->log("url");
        $this->log($url);
        $this->log("fields");
        $this->log($fields);
        curl_close($ch);
        if ($result)
            return $result;
        else
            return false;
    }

    private function log($message) {
        print_r($message);
        echo "\n";
    }

}

扩展 从核心文件扩展 Admin_controller。

我的 Controller 代码是:

<?php
defined('BASEPATH') OR exit('No direct script access allowed');
class Myoperator extends Admin_controller
{

    public function __construct()
    {
        parent::__construct();
        //$this->load->model('myoperator_model');
    }

    public function index()
   {
       try {
           $this->load->library('my_operator');  
           $data = $this->my_operator->run();
       } catch (Exception $e) {
           var_dump($e->getMessage());
       }

       $this->load->view('admin/myoperator/view', $data);
   }
}

我的 View 代码是:

<?php init_head(); ?>
    <div id="wrapper">
        <div>
            <?php
                echo $this->MY_Operator->run();
            ?>
        </div>
    </div>
<?php init_tail(); ?>

</body>
</html>

我不会得到想要的输出。 为此，我引用了:

codeigniter data passing controller->library->view

Pass data from library to controller and then to view in CodeIgniter 2

passing parameter from view to library and return after process Code igniter

当我运行这段代码时出现错误

This page isn’t working localhost is currently unable to handle this request. HTTP ERROR 500

根据我的理解，如果服务器配置有任何问题，就会出现此错误，但同一应用程序的其他页面运行完美。

已编辑: 以前的错误已解决，一些链接问题，但是当我扩展该类时，出现新错误:

Unable to locate the specified class: Session.php

是否可以将核心文件类扩展为库类？

请帮我解决这个问题。任何形式的帮助表示赞赏。提前致谢。

Answer 1

解决了我的问题，只是忘了在 Myoperator Controller 文件中包含 Admin_controller 文件。

这是我更新的代码:

<?php
defined('BASEPATH') OR exit('No direct script access allowed');
require APPPATH. 'core/Admin_controller.php'; //forgot this line of code
class Myoperator extends Admin_controller
{

    public function __construct()
    {
        parent::__construct();
        //$this->load->model('myoperator_model');
    }

    public function index()
   {
       try {
           $this->load->library('my_operator');  
           $data = $this->my_operator->run();
       } catch (Exception $e) {
           var_dump($e->getMessage());
       }

       $this->load->view('admin/myoperator/view', $data);
   }
}