我正在尝试使用PHP Data API将视频上传到YouTube
$yt = new Zend_Gdata_YouTube($httpClient);
$myVideoEntry = new Zend_Gdata_YouTube_VideoEntry();
$filesource = $yt->newMediaFileSource('mytestmovie.mov');
$filesource->setContentType('video/quicktime');
$filesource->setSlug('mytestmovie.mov');
$myVideoEntry->setMediaSource($filesource);
$myVideoEntry->setVideoTitle('My Test Movie');
$myVideoEntry->setVideoDescription('My Test Movie');
// Note that category must be a valid YouTube category !
$myVideoEntry->setVideoCategory('Comedy');
// Set keywords, note that this must be a comma separated string
// and that each keyword cannot contain whitespace
$myVideoEntry->SetVideoTags('cars, funny');
// Optionally set some developer tags
$myVideoEntry->setVideoDeveloperTags(array('mydevelopertag',
'anotherdevelopertag'));
// Optionally set the video's location
$yt->registerPackage('Zend_Gdata_Geo');
$yt->registerPackage('Zend_Gdata_Geo_Extension');
$where = $yt->newGeoRssWhere();
$position = $yt->newGmlPos('37.0 -122.0');
$where->point = $yt->newGmlPoint($position);
$myVideoEntry->setWhere($where);
// Upload URI for the currently authenticated user
$uploadUrl =
'http://uploads.gdata.youtube.com/feeds/users/default/uploads';
// Try to upload the video, catching a Zend_Gdata_App_HttpException
// if availableor just a regular Zend_Gdata_App_Exception
try {
$newEntry = $yt->insertEntry($myVideoEntry,
$uploadUrl,
'Zend_Gdata_YouTube_VideoEntry');
} catch (Zend_Gdata_App_HttpException $httpException) {
echo $httpException->getRawResponseBody();
} catch (Zend_Gdata_App_Exception $e) {
echo $e->getMessage();
}
有谁知道如何从$ newEntry对象获取上传视频的网址。
任何帮助将不胜感激:)
答案 0 :(得分:4)
试试这个:
try {
$newEntry = $yt->insertEntry($myVideoEntry, $uploadUrl,
'Zend_Gdata_YouTube_VideoEntry');
$id = $newEntry->getVideoId(); // YOUR ANSWER IS HERE :)
echo $id;
}