Cakephp:如何从Google应用程序引擎检索视频url

Cakephp:如何从Google应用程序引擎检索视频url,php,google-app-engine,cakephp,Php,Google App Engine,Cakephp,上传视频: require_once 'google/appengine/api/cloud_storage/CloudStorageTools.php'; use google\appengine\api\cloud_storage\CloudStorageTools; $gs_name = $file1['tmp_name']; move_uploaded_file($gs_name, 'gs://bucketname/videos/' . $file_name); 获取视频: $f

上传视频

require_once 'google/appengine/api/cloud_storage/CloudStorageTools.php';  
use google\appengine\api\cloud_storage\CloudStorageTools;

$gs_name = $file1['tmp_name'];
move_uploaded_file($gs_name, 'gs://bucketname/videos/' . $file_name);
获取视频:

$file_url = CloudStorageTools::getPublicUrl("gs://bucketname/videos/20150225074324830.mp4", true);
我无法获取视频,请告诉我如何从Google App Engine获取视频。通过以下方式成功从google app engine获取图像:

$image_url = CloudStorageTools::getImageServingUrl('gs://bucketname/avatars/' . $user['User']['id'] . '/' . $user['User']['avatar']);  

我解决了问题,获取视频的代码是:

$file_url = CloudStorageTools::getPublicUrl("gs://bucketname/videos/20150225074324830.mp4", true);
问题是我必须通过编写以下代码将视频标记为公共视频:

$storeAt="gs://bucketname/videos/20150225074324830.mp4";
$options = array('gs'=>array('acl'=>'public-read','Content-Type' =>   $file['type']));
$ctx = stream_context_create($options);
$saved = file_put_contents($storeAt, file_get_contents($file['tmp_name']), 0, $ctx);