Php 我可以在codeigniter中的函数之间传递数据吗？

我正在加载一个简单的表单作为子视图，我需要将数据传递给下一个函数，但是当我使用

$this->data

时，我会得到一个空白页面。这是我的控制器：

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

class Search extends Admin_Controller {


    function __construct()
    {
        parent::__construct();
        $this->load->library('form_validation');
        $this->load->helper('form');
        $this->load->model('search_model');

    }
    function index()
    {

        $this->data['subview'] = ('../views/user/search_form');
        $this->load->view('../views/_layout_admin', $this->data);
    }
    function execute_search()
    {
        // Retrieve the posted search term.
        $search_term = $this->input->post('search');

        // Use a model to retrieve the results.
        $data['results'] = $this->search_model->get_results($search_term);

        // Pass the results to the view.
        $this->data['subview'] = ('../views/user/search_results');
        $this->load->view('../views/user/search_results', $this->data);

    }
    }

---

您也可以通过在主视图中执行以下操作加载子视图：

<?php $this->load->view('user/search_form'); ?>

用以下代码替换控制器：

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

class Search extends Admin_Controller {


    function __construct()
    {
        parent::__construct();
        $this->load->library('form_validation');
        $this->load->helper('form');
        $this->load->model('search_model');

    }
    function index()
    {

        $data['subview'] = ('../views/user/search_form');
        $this->load->view('layout_admin', $data);
    }
    function execute_search()
    {
        // Retrieve the posted search term.
        $search_term = $this->input->post('search');

        // Use a model to retrieve the results.
        $data['results'] = $this->search_model->get_results($search_term);

        // Pass the results to the view.
        $data['subview'] = ('../views/user/search_results');
        $this->load->view('search_results', $data);

    }
    }

您好，我认为您需要阅读所有codeigniter的文档，因为当您在
应用程序/views/…
中加载视图时，您不需要使用
$this->load->view（'../views/--
仅
$this->load->view（'views/…
然后，当您向视图发送$data时，您需要的是
$this->load->view（'../views/view\u-tamplate'，$data）；
作为文档，您使用$this->data的目的是什么？为什么不使用$data？