通过url id验证youtube视频

问题描述：

I am using php and yt API to determine if a video exists. I have implemented two functions two help me along this cause. The issues is with the function isYoutubeVideo is returning me null value when a video is valid or invalid. I have checked its paremeters but I am still not sure why is giving me null value. Do i have the isYoutubeVideo function set up wrong/missing something/needs change?

        function isYoutubeVideo($formatted_url) {
            $isValid = false;
            if (isValidURL($formatted_url)) {
                $idLength = 11;
                $idStarts = strpos($formatted_url, "?v=");
                if ($idStarts === FALSE) {
                    $idStarts = strpos($formatted_url, "&v=");
                }
                if ($idStarts !== FALSE) {
                    //there is a videoID present, now validate it
                    $v = substr($formatted_url, $idStarts, $idLength);
                    $headers = get_headers('http://gdata.youtube.com/feeds/api/videos/' . $videoID);
                    //did the request return a http code of 2xx?
                    if (!strpos($headers[0], '200')) {
                        $isValid = true;
                    }
                }
            }
            return $isValid;

我使用php和yt API来确定视频是否存在。我已经实现了两个功能，两个帮助我这个原因。问题是函数isYoutubeVideo code>在视频有效或无效时返回null值。我检查了它的参数，但我仍然不确定为什么给我空值。我是否有 isYoutubeVideo code>功能设置错误/缺少某些/需要更改？ p>

  function isYoutubeVideo（$ formatted_url）{
 $ isValid = false; 
 if（isValidURL（$ formatted_url））{
 $ idLength = 11; 
 $ idStarts  = strpos（$ formatted_url，“？v =”）; 
 if（$ idStarts === FALSE）{
 $ idStarts = strpos（$ formatted_url，“＆amp; v =”）; 
} 
 if（n if）  $ idStarts！== FALSE）{
 //存在一个videoID，现在验证它
 $ v = substr（$ formatted_url，$ idStarts，$ idLength）; 
 $ headers = get_headers（'http：//  gdata.youtube.com/feeds/api/videos/'。$ videoID）; 
 //请求返回的http代码为2xx？
 if（！strpos（$ headers [0]，'200'））  {
 $ isValid = true; 
} 
} 
} 
返回$ isValid; 
  code>  pre> 
  div>

答

Why aren't your first two echo statements also echoing <span id="resultval"> ?;

With the current markup and jQuery code this is why jQuery populates #special with null.

You need to change you PHP code to:

echo('<div id="special"><span id="resultval">The YouTube video does not exist.</span></div>');

echo('<div id="special"><span id="resultval">that is a valid youtube video</span></div>');

Although, I do not understand why you used such a roundabout way to populate #special. You could have just started with a <p> inside <div id="#special"> and then use the selector $("#special > p")

Good luck!

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