问题
I'm creating a website using symfony for blogging. Users can upload their posts to the site. when a user add a file it will save inside web/upload/file_upload
and the file path will save in add_post
table. When a admin view a add_post
table template he can see the path of the downloaded file of each and every user, what i want to do is through this file path download the file.
How can i do this?
edit 1:
Model - Blog_user Module - post
Table Structre - table name- Blog_user
1 user_id bigint(20)
2 gender varchar(255)
3 blog_status tinyint(1)
4 file varchar(255)
Form
'user_id' => new sfWidgetFormInputHidden(),
'gender' => new sfWidgetFormInputText(),
'file' => new sfWidgetFormInputFile(),
Here when uploading a file, the filepath save in Blog_user table and file save inside web/upload directory.
edit 2:
//save file method
public function saveFile(){
$file = $this->getValue('file');
if(isset($file)){
$filename = 'POST_Uploaded -' .($file->getOriginalName());
$file->save(sfConfig::get('sf_upload_dir').'/post_upload'.'/'.$filename);
}
}
E:\xampp\htdocs\trunk\web\uploads\post_upload\POSt_Uploaded -JS.pdf
This how it saved in side web/upload/post_upload directory and same path will save inside db also
edit 3:
when a user upload a blog it will save in blog_user table and it consist blog _id as primary key, user_id is on user table. what i want do is when user upload a file , both user_id , and blog_id should be saved inside blog table . how to do it?
user table - user_id , file(uploaded file)
blog table - blog_id - there are blog titles , each title has an unique blog id , so that user can upload a file under each titles,
post table - post_id, blog_id, user_id
回答1:
Assuming:
- your module name is
moduleName
- the model with the file is
BlogUser
- the primary key of the model is
id
I will go this way:
In your template:
<a href="<?php echo url_for('post/download?user_id='.$blog_user->getUserId()) ?>">Download file</a>
Then, in your action (use the function from Miqdad Ali):
public function executeDownload(sfwebRequest $request)
{
$blog_user = Doctrine_Core::getTable('Blog_user')->find($request->getParameter('user_id'));
$this->forward404Unless($blog_user);
header('content-type:');
header('Content-Description: File Transfer');
//header('Content-Type: application/octet-stream');
header('Content-Disposition: attachment; filename='.basename($blog_user->getFile()));
header('Content-Transfer-Encoding: binary');
header('Expires: 0');
header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
header('Pragma: public');
header('Content-Length: ' . filesize($blog_user->getFile()));
ob_clean();
flush();
readfile($blog_user->getFile());
return sfView::NONE;
}
回答2:
You can write this file inside any controller and call that function while user clicking on the download
function download(){
$file = "path/to/the/file.zip";
if (file_exists($file)) {
exit;
}
header('content-type:');
header('Content-Description: File Transfer');
//header('Content-Type: application/octet-stream');
header('Content-Disposition: attachment; filename='.basename($file));
header('Content-Transfer-Encoding: binary');
header('Expires: 0');
header('Cache-Control: must-revalidate, post-check=0, pre-check=0');
header('Pragma: public');
header('Content-Length: ' . filesize($file));
ob_clean();
flush();
readfile($file);
exit;
}
来源:https://stackoverflow.com/questions/10894166/how-to-download-a-file-on-clicking-thefile-path-using-php-symfony