我可以使用 xml 从 youtube API 获取页面令牌吗?

Question

我正在开发youtube api并开发一个youtube api网站，我想获取视频列表的下一页令牌，但我正在从xml提要获取数据。

如何从 xml 源获取下一页令牌？

这是我的代码：

<?php
// set feed URL
$feedURL = 'http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed';
// read feed into SimpleXML object
$sxml = simplexml_load_file($feedURL);
?>
<h1><?php echo $sxml->title; ?></h1>
<?php
// iterate over entries in feed
foreach ($sxml->entry as $entry) {
  // get nodes in media: namespace for media information
  $media = $entry->children('http://search.yahoo.com/mrss/');
  // get video player URL
  $attrs = $media->group->player->attributes();
  $watch = $attrs['url']; 
  // get video thumbnail
  $attrs = $media->group->thumbnail[0]->attributes();
  $thumbnail = $attrs['url']; 
  // get <yt:duration> node for video length
  $yt = $media->children('http://gdata.youtube.com/schemas/2007');
  $attrs = $yt->duration->attributes();
  $length = $attrs['seconds']; 
  // get <yt:stats> node for viewer statistics
  $yt = $entry->children('http://gdata.youtube.com/schemas/2007');
  $attrs = $yt->statistics->attributes();
  $viewCount = $attrs['viewCount']; 
  // get <gd:rating> node for video ratings
  $gd = $entry->children('http://schemas.google.com/g/2005'); 
  if ($gd->rating) {
    $attrs = $gd->rating->attributes();
    $rating = $attrs['average']; 
  } else {
    $rating = 0; 
  }
  $my_format = 'mqdefault.jpg'; 
  $my_url = 'img.ytapi.com';
  ?>
  <div class="item">
    <span class="title">
      <a href="<?php echo $watch; ?>"><?php echo $media->group->title; ?></a>
    </span>
    <p>
      <span class="thumbnail">
        <a href="<?php echo $watch; ?>"><img src="<?php echo str_replace(array("i.ytimg.com","0.jpg"),array($my_url, $my_format), $thumbnail);?>" /></a>
        <br/>click to view
      </span> 
      <span class="attr">By:</span> <?php echo $entry->author->name; ?> <br/>
      <span class="attr">Duration:</span> <?php printf('%0.2f', $length/60); ?> 
      min. <br/>
      <span class="attr">Views:</span> <?php echo $viewCount; ?> <br/>
      <span class="attr">Rating:</span> <?php echo $rating; ?> 
    </p>
  </div>      
<?php
}
?>

Answer 1

您可以使用

start-index 参数请求其他页面。此参数与max-results参数结合使用。所以：

如果省略 max-results 参数，则请求以下 URL：

• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed结果 1-25
• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed?start-index=26结果 26-50
• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed?start-index=51结果 51-75

如果设置了max-results=10则请求以下 URL：

• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed?max-results=10结果 1-10
• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed?max-results=10&start-index=11结果 11-20
• 
  http://gdata.youtube.com/feeds/api/standardfeeds/most_viewed?max-results=10&start-index=21结果 21-30