Question

我是CodeIgniter和HMVC框架的新手，我正在创建一个示例测试图像下拉列表，以获取有关如何解决本周应完成的大型项目的蓝图。我的问题是，如何使用下拉列表以表格式搜索和显示过滤后的数据库值？例如，如果我选择下拉列表中的值之一并单击“搜索”，则所有检索到的值将被过滤并根据下拉列表中选择的值显示。这是图像下拉菜单的示例测试代码的代码：

这是我模型中的代码：

class Image_model extends CI_Model{

    function __construct(){
        parent::__construct();
    }

    /*Sample test model function for the image dropdown list*/

    //display the images table
    public function displayTableImages()
    {
        $query = $this->db->select('main_image, main_image_url');
        $query = $this->db->from('product_master');
        $query = $this->db->get();
        return $query->result();
    }

    //dropdown search list for images
    public function searchDropdownImages($type)
    {
        switch ($type) {

            case 'all':
                {
                    $query = $this->db->where("main_image != '', 'main_image_url != ''");
                }
            case 'with-image':
                {
                    $query = $this->db->where("'main_image != '', 'main_image_url != ''");
                }
            break;
            case 'no-image':
                {
                    $query = $this->db->where("'main_image = ''",  "'main_image_url = ''");
                }   
            break;
            default:
                    $query = $this->db->select('main_image', 'main_image_url');
                    $query = $this->db->from('product_master');
                break;
                    if ($query->num_rows()) {
                    return $query->result();
                    }
            }
    }
    /*End sample test function*/
}

这是我的控制器的代码：

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

class Sample_image_dropdown extends MX_Controller{

    public function __construct()
    {
        parent::__construct();
    }

    public function index()
    {
        $data['main_view'] = 'sample_view/image_dropdown_view';
        $this->display_table_images();
    }

    public function display_table_images()
    {
        $this->load->model('Sample_model/image_model');
        $data['images'] = $this->image_model->displayTableImages();
        $data['main_view'] = 'sample_view/image_dropdown_view';
        $this->load->view('sample_view/image_dropdown_view', $data);
    }

    public function search_dropdown_images($type)
    {
        $this->load->model('Sample_model/image_model');

        $type['dropdown_images'] = $this->input->post('type');
        switch ($type) {
            case 'all':
                $data['images'] = $this->image_model->get('all');                   
                break;
            case 'with-image':
                $data['images'] = $this->image_model->get('with-image');
                break;
            case 'no-image':
                $data['images'] = $this->image_model->get('no-image');
                break;
            default:
                echo 'There are no images to be returned';
                break;
        }
        $this->load->view('sample_view/image_dropdown_view', $type);
    }
}

这是我认为的代码：

<!DOCTYPE html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <title>Document</title>
</head>
<body>
    <div>
            <div>
                <?php echo "<select name='type' id='type'>
                                <option value='all'>All Image</option>
                                <option value='with-image'>With Image</option>
                                <option value='no-image'>Without Image</option>
                            </select>" ?>

                <?php echo "<input type='submit' value='Search'>"; ?>
            </div>
            <div>
                <h3>Images</h3>
            </div>
            <div>
                <?php if (isset($images)): ?>
                <?php foreach($images as $image): ?>
                    <table border="1">
                        <tr>
                            <td><?php echo "$image->main_image"; ?></td>
                            <td><?php echo "$image->main_image_url"; ?></td>
                        </tr>
                    </table>
                <?php endforeach; ?>
                <?php endif; ?>
            </div>
    </div>
</body>
</html>

对于您可以就我的问题提供的任何帮助/建议，我将非常高兴。

Answer 1

您应该为此使用AJAX。首先，您必须将一个查询从ajax发送到调用控制器，然后使用该控制器从数据库加载数据，并将其作为HTML块发送到视图并显示它。并在链接到ajax的下拉列表中创建一个OnChange事件希望这可以帮助您

Answer 2

使用分页库和get方法来过滤数据... get方法将帮助您创建base_url。您需要做的就是在where语句中使用get数据。您面临的问题是，当您转到另一个页面时，您丢失了所选的选项。在控制器中写入孔后，只需更改base_url以配置分页并将其设置为

if(isset($_GET))
            {
                $config['first_url']        = $config['base_url'].'/1'.'?'.http_build_query($_GET, '', "&");    
            }

这将有助于生成正确的链接并从您的角度来看..获得所选选项的所选值。

<select name='type' id='type'>
    <option value='all' <?php echo (isset($_get['type']) && $_get['type'] == 'with-image')?'selected':'';  ?> >
      All Image
    </option>
    <option value='with-image'  <?php echo (isset($_get['type']) && $_get['type'] == 'with-image')?'selected':'';  ?>  >
      With Image
    </option>
    <option value='no-image'  <?php echo (isset($_get['type']) && $_get['type'] == 'no-image')?'selected':'';  ?>  >
       Without Image
    </option>
</select>

如何使用CodeIgniter中的下拉列表在HTML表中显示过滤的数据库值？

2 个答案: