code.google.com的API为PHP让不同的视频ID的YouTube
我已经使用这个功能:功能printVideoEntry($的VideoEntry),并得到了
Video ID: XXXXXXXXXXXXXXXXXXXXXXXXXX
Watch page: http://www.youtube.com/watch?v=YYYYYYYYYYYYYY&feature=youtube_gdata_player
Flash Player Url: http://www.youtube.com/v/YYYYYYYYYYYYY?version=3&f=playlists&app=youtube_gdata
如果我通过给定的视频ID它给了我错误:
Uncaught exception 'Zend_Gdata_App_HttpException' with message 'Expected response code 200, got 400 Invalid id'
如果我从观看网页和Flash播放器网址YYYYYYYYYYYYYYYYY通[两者都是相同的]我得到它我需要什么。
帮助非常感谢,提前致谢。
使用此功能为获得参赛视频
function printVideoEntry($videoEntry) {
echo "<div onclick=\"ytvbp.presentVideo('".$videoEntry->getVideoId()."')\" >";
echo 'Video: '.$videoEntry->getVideoTitle() . "<br>";
echo "</div>";
echo 'Video ID: ' . $videoEntry->getVideoId() . "<br>";
echo 'Watch page: ' . $videoEntry->getVideoWatchPageUrl() . "<br>";
echo 'Flash Player Url: ' . $videoEntry->getFlashPlayerUrl() . "<br>";
}
我打电话打印视频功能从
function getAndPrintPlaylistVideoFeed($playlistListEntry) {
$yt = new Zend_Gdata_YouTube();
$playlistVideoFeed = $yt->getPlaylistVideoFeed($playlistListEntry->getPlaylistVideoFeedUrl());
foreach ($playlistVideoFeed as $playlistVideoEntry) {
$getandprintplaylistvideofeed_array[] = printVideoEntry($playlistVideoEntry);
}
道歉但这是我在stackoverflow上的第一个问题。 – 2012-02-01 14:51:38