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问题描述
我使用PHP来访问YouTube视频的属性,例如标题,URL,缩略图,持续时间等。
I'm using PHP to access YouTube video attributes e.g. title, url, thumbnail, duration etc.
这并不是说GD:等级(5星级)已取代YT:等级(喜欢/不喜欢系统),我想修改我的PHP
Not that gd:rating (5 star ratings) has been replaced by yt:rating (like/dislike system), I'm trying to modify my PHP.
GD:评级(旧)
$gd = $entry->children('http://schemas.google.com/g/2005');
if ($gd->rating) {
$attrs = $gd->rating->attributes();
$rating = $attrs['average'];
} else {
$rating = 0;
}
YT:等级(新)
$yt = $entry->children('http://gdata.youtube.com/schemas/2007');
if ($yt->rating && $yt->rating[0]->attributes()) {
$attrs = $yt->rating[0]->attributes();
$videoobj[$loopCounter]['dislikes'] = strval($attrs['numDislikes']);
$videoobj[$loopCounter]['likes'] = strval($attrs['numLikes']);
} else {
$videoobj[$loopCounter]['dislikes'] = 0;
$videoobj[$loopCounter]['likes'] = 0;
}
使用YT:评级code没有工作。即使我做了的print_r($ YT->评级);
,没有什么数组中
我在哪里去了?
推荐答案
校正。就可以了。
您访问的需求来追加到结尾的网址:
The URL you access needs this appended to the end:
$url = 'http://gdata.youtube.com/feeds/api/videos/' . $vid . '?v=2';
这一切都不同 - 所以我的code最后一点:
That made all the difference - so the last bit of my code:
// get <yt:rating> node for like/dislikes
$yt = $entry->children('http://gdata.youtube.com/schemas/2007');
$attrs = $yt->rating->attributes();
$obj->dislikes = $attrs['numDislikes'];
$obj->likes = $attrs['numLikes'];
echo $obj->dislikes;
echo ' ';
echo $obj->likes;
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