[英]How to display the values of database table in CodeIgniter?
我正在創建一個示例圖像測試,其中我想在 CodeIgniter 的數據庫表中顯示所有圖像。 問題是,如何使用 CodeIgniter 顯示數據庫表的值?
這是我的模型的代碼:
<?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*/
}
?>
這是我的控制器的代碼:
<?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', $data);
}
}
?>
這是我的觀點的代碼:
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Document</title>
</head>
<body>
<div>
<select name="" id="">
<option value="all">All Image</option>
<option value="with-image">With Image</option>
<option value="no-image">Without Image</option>
</select>
<input type="submit" value="Search">
</div>
<?php if (isset($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>
您在模型和控制器中犯了一個錯誤,我提到了更改,請記下並嘗試,
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', $data);
}
我假設您在uploads/abc.png
時將圖像存儲在目錄中uploads/abc.png
然后你可以從db獲取數據(你已經在做)
並認為:
<img src="echo dir_path./$image->main_image_url;" />
或者,如果您將整個路徑存儲在DB 中,則:
<img src="echo $image->main_image_url;" />
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