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，以获得如何解决本周到期的大项目的蓝图。我的问题是，如何使用下拉列表以表格格式搜索和显示已过滤的数据库值？例如，如果我选择下拉列表中的一个值并单击搜索，则将过滤所有检索到的值，并根据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', $type);
}
}
这是我视图中的代码：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
</head>
<body>
<div>
<div>
<?php echo "<select name='type' id='type'>
<option value='all'>All Image</option>
<option value='with-image'>With Image</option>
<option value='no-image'>Without Image</option>
</select>" ?>
<?php echo "<input type='submit' value='Search'>"; ?>
</div>
<div>
<h3>Images</h3>
</div>
<div>
<?php if (isset($images)): ?>
<?php foreach($images as $image): ?>
<table border="1">
<tr>
<td><?php echo "$image->main_image"; ?></td>
<td><?php echo "$image->main_image_url"; ?></td>
</tr>
</table>
<?php endforeach; ?>
<?php endif; ?>
</div>
</div>
</body>
</html>
我很乐意感谢您就我的问题提供的任何帮助/建议。
你应该使用AJAX。首先，您必须从ajax向调用控制器发送查询，然后使用控制器从数据库加载数据并将其作为HTML块发送到视图并显示它。并将一个OnChange事件发送到与ajax链接的下拉列表希望这个可以帮助你
使用分页库和get方法来过滤数据... get方法将帮助您创建base_url。您需要做的就是将where数据用于where语句。您遇到的问题是当您转到另一个页面时，您丢失了所选的选项。在写入孔之后的控制器中只需更改base_url以进行配置分页并将其设置为
if(isset($_GET))
{
$config['first_url'] = $config['base_url'].'/1'.'?'.http_build_query($_GET, '', "&");
}
它将有助于生成正确的链接，并在您的视图侧..获取所选选项的选定值..
<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>