如何使用codeigniter中的下拉列表在html表中显示过滤后的数据库值?

qni6mghb  于 2021-06-17  发布在  Mysql
关注(0)|答案(2)|浏览(302)

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,以便得到一个关于如何解决这个大项目的蓝图,这个项目我将在本周完成。我的问题是,如何使用下拉列表以表格格式搜索和显示过滤后的数据库值?例如,如果我在下拉列表中选择一个值并单击“搜索”,则所有检索到的值都将被过滤,并将根据在下拉列表中选择的值显示。以下是图像下拉列表的示例测试代码:
以下是我的模型中的代码:

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*/
}

以下是我的控制器代码:

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);
    }
}

以下是我认为的代码:

<!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>

如果您能就我的问题提供任何帮助/建议,我将不胜感激。

s4chpxco

s4chpxco1#

使用分页库和get方法筛选数据。。。get方法将帮助您创建基本url。您所需要做的就是使用GETDATA in FORWHERE语句。您面临的问题是,当您转到另一页时,您丢失了所选的选项。在控制器中,在写完洞之后,只需更改基本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, '', "&");    
            }

这将有助于生成适当的链接,并在您的观点方面。。获取选定选项的选定值。。

<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>
balp4ylt

balp4ylt2#

你应该使用ajax。首先,您必须从ajax向callcontroller发送一个查询,然后使用controller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,希望这一个可以帮助您

相关问题