我正在创建一个示例图像测试，其中我想在CodeIgniter中显示数据库表中的所有图像。问题是，如何使用CodeIgniter显示数据库表的值？
这是我的模型的代码：
<?php
class Image_model extends CI_Model{
/*Sample test function for the image dropdown list*/
public function getImages()
{
$query = $this->db->select('main_image_url'));
$this->db->get('product_master');
return $query->result();
}
/*End sample test function*/
}
?>
这是我的控制器的代码：
<?php
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->load->view('sample_view/image_dropdown_view', $data);
}
public function display_all_images()
{
$this->load->model('sample_model/image_model');
$data['images'] = $this->image_model->getImages();
$data['main_view'] = "sample_view/image_dropdown_view";
$this->load->view('sample_view/image_dropdown_view', $data);
}
}
?>
以下是我的观点代码：
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
</head>
<body>
<div>
<select name="" id="">
<option value="all">All Image</option>
<option value="with-image">With Image</option>
<option value="no-image">Without Image</option>
</select>
<input type="submit" value="Search">
</div>
<?php if (isset($images)):?>
<?php foreach ($images as $image):?>
<table>
<tr>
<td>
<th>
Images
</th>
</td>
</tr>
<tr>
<td>
<img src="<?php echo "$image->main_image_url";?>">
</td>
</tr>
</table>
<?php endforeach; ?>
<?php endif; ?>
</body>
</html>
你在你的模型和控制器中犯了一个错误，我提到了变化，请记下来试试，
模型文件
class Image_model extends CI_Model{
/*Sample test function for the image dropdown list*/
public function getImages()
{
$query = $this->db->select('main_image_url');
$this->db->from('product_master');
$query = $this->db->get();
return $query->result();
}
/*End sample test function*/
}
控制器文件，
public function __construct()
{
parent::__construct();
}
public function index()
{
$data['main_view'] = 'sample_view/image_dropdown_view';
$this->display_all_images();
}
public function display_all_images()
{
$this->load->model('sample_model/image_model');
$data['images'] = $this->image_model->getImages();
$data['main_view'] = "sample_view/image_dropdown_view";
$this->load->view('sample_view/image_dropdown_view', $data);
}
我假设您在uploads/abc.png上传时将图像存储在目录中
然后你可以从db（你已经在做）中获取数据
在视野中：
<img src="echo dir_path./$image->main_image_url;" />
或者如果要将整个路径存储在DB中，则：
<img src="echo $image->main_image_url;" />