Getting different video ids YouTube - php

Question

code.google.com's API for PHP
I have used this function : function printVideoEntry($videoEntry) and got
Video ID: XXXXXXXXXXXXXXXXXXXXXXXXXX Watch page:
http://www.youtube.com/watch?v=YYYYYYYYYYYYYY&feature=youtube_gdata_player
Flash Player Url:
http://www.youtube.com/v/YYYYYYYYYYYYY?version=3&f=playlists&app=youtube_gdata
If I pass given video id it gives me error:
Uncaught exception 'Zend_Gdata_App_HttpException' with message
'Expected response code 200, got 400 Invalid id'
And if I pass YYYYYYYYYYYYYYYYY from Watch page and Flash Player Url [both are same] I'm getting it what I need.
Help is much appreciated, Thanks in advance.
Using this function for getting video entries
function printVideoEntry($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: ' . $videoEntry->getFlashPlayerUrl() . "<br>";
}
I'm calling print video function from
function getAndPrintPlaylistVideoFeed($playlistListEntry) {
$yt = new Zend_Gdata_YouTube();
$playlistVideoFeed = $yt->getPlaylistVideoFeed($playlistListEntry->getPlaylistVideoFeedUrl());
foreach ($playlistVideoFeed as $playlistVideoEntry) {
$getandprintplaylistvideofeed_array[] = printVideoEntry($playlistVideoEntry);
}

Answer 1

$videoEntry should be a VideoEntry-object, which you can get based on an ID:
$videoEntry = $yt->getVideoEntry('the0KZLEacs');
Then you could call the function and set that as a parameter:
printVideoEntry($videoEntry);
Or you could edit the first lines of printVideoEntry:
function printVideoEntry($videoId) {
$videoEntry = $yt->getVideoEntry('the0KZLEacs');
//...
}
Check out the docs to see what information you can squeeze out of a VideoEntry =)