Question

我将一个简单的表单作为子视图加载，我需要将数据传递给下一个函数，但是当我使用$this->data时，我得到一个空白页面。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);

    }
    }

任何帮助/想法？非常贬值

Answer 1

您也可以在主视图中加载子视图：

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

Answer 2

使用以下代码替换您的控制器：

 <?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()
    {

        $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);

    }
    }

我可以在codeigniter中的函数之间传递数据吗？

2 个答案: