此圖片顯示了我需要笨做。我有一個頁面有幾個div標籤。我需要上傳圖片並在相同的地方顯示。但是應該有3個不同的圖像和3個不同的文件位置來保存這些圖像。我嘗試了很多方法。請有想法的人幫助我。
我控制器
<?php
if (!defined('BASEPATH'))
exit('No direct script access allowed');
class Upload_Controller extends CI_Controller
{
public function __construct()
{
parent::__construct();
}
public function index(){
$this->load->view('file_view', array(
'error' => ' '
));
}
public function file_view()
{
$this->load->view('file_view', array(
'error' => ' '
));
}
public function do_upload()
{
$config = array(
'upload_path' => "./uploads/",
'allowed_types' => "gif|jpg|png|jpeg|pdf",
'overwrite' => TRUE,
'max_size' => "2048000", // Can be set to particular file size , here it is 2 MB(2048 Kb)
'max_height' => "768",
'max_width' => "1024"
);
$this->load->library('upload', $config);
if ($this->upload->do_upload()) {
$data = array(
'upload_data' => $this->upload->data()
);
$this->load->view('file_view', $data);
} else {
$error = array(
'error' => $this->upload->display_errors()
);
$this->load->view('file_view', $error);
}
}
public function do_upload2()
{
$config = array(
'upload_path' => "./uploads/index2/",
'allowed_types' => "gif|jpg|png|jpeg|pdf",
'overwrite' => TRUE,
'max_size' => "2048000", // Can be set to particular file size , here it is 2 MB(2048 Kb)
'max_height' => "768",
'max_width' => "1024"
);
$this->load->library('upload', $config);
if ($this->upload->do_upload()) {
$data = array(
'upload_data' => $this->upload->data()
);
$this->load->view('file_view', $data);
} else {
$error = array(
'error1' => $this->upload->display_errors()
);
$this->load->view('file_view', $error);
}
}
}
?>
我查看
<div id="1">
<?php echo form_open_multipart('upload_controller/do_upload');?>
<?php echo "<input type='file' name='userfile' size='20' />"; ?>
<?php echo "<input type='submit' name='submit' value='upload' /> ";?>
<?php echo "</form>"?>
</div>
<div id="2">
<h3>Your file was successfully uploaded!</h3>
<!-- Uploaded file specification will show up here -->
<ul>
<li>
</li>
<img alt="Your uploaded image" src="<?=base_url(). 'uploads/' . $upload_data['file_name'];?>">
</ul>
</div>
<div id="3">
<?php echo form_open_multipart('upload_controller/do_upload2');?>
<?php echo "<input type='file' name='userfile' size='20' />"; ?>
<?php echo "<input type='submit' name='submit' value='upload' /> ";?>
<?php echo "</form>"?>
</div>
<div id="4">
<h3>Your file was successfully uploaded!</h3>
<!-- Uploaded file specification will show up here -->
<ul>
<li>
</li>
<img alt="Your uploaded image" src="<?=base_url(). 'uploads/index2/' . $upload_data['file_name'];?>">
</ul>
<p>
<?php echo anchor('upload_controller/file_view', 'Upload Another File!'); ?>
</p>
</div>
您的問題太寬泛。你究竟需要回答什麼?請解釋你嘗試過的方法。 – josephting
我將上面的代碼添加到Mr.josephting – Dushee