I want to download 4 different files through 4 different links. I am using the Media view to download the files, but I have to hardcode the file name in the download functions in the controller:
function download () {
$this->view = 'Media';
$params = array(
'id' => 'example.zip',
'name' => 'example',
'download' => true,
'extension' => 'zip',
'path' => APP . 'files' . DS
);
$this->set($params);
}
This works fine for one file. Now, for links number 2,3,4, do I need to create 3 different actions and give different file names in them, or is there a way in which I can use download() to only 开发者_StackOverflow中文版download the respective file depending on which link has been clicked?
That's what variables are for. Generic example:
function download($fileId) {
$file = // find the file you want to serve based on $fileId
$pathInfo = pathinfo($file['path']);
$this->view = 'Media';
$params = array(
'id' => $file['name'],
'name' => $pathInfo['filename'],
'extension' => $pathInfo['extension'],
'download' => true,
'path' => APP . 'files' . DS
);
$this->set($params);
}
In CakePhp 2.x you will get error The view for YourController::download() was not found.
Use viewClass field in CakePHP 2.x:
$this->viewClass = 'Media';
See Media Views — CakePHP Cookbook v2.x documentation
UPD: Media Views are deprecated since CakePHP 2.3, and CakeResponse::file()
should be used:
$this->response->file($file['path'], array('download' => true, 'name' => 'foo'));
return $this->response;
精彩评论