video - 如果视频是 “unavailable”，是否有修复程序可跳过播放列表中的YouTube视频

Question

我在HTML 5媒体播放器播放列表中有YouTube视频的汇总列表(从mySql数据库查询)。但是随着时间的流逝，由于版权或判断问题，YouTube将禁用某些视频，但是这些链接仍在我的列表中。任何人都可以推荐JS或其他解决方案或文章，以查看视频链接是否在x时间内开始启动跳过或下一步操作。请指教。

没有解决方案，因为广泛的谷歌搜索没有建议。

我的基本逻辑是，如果x秒钟后视频无法播放，则跳过，否则播放。

// THIS ACTUALLY CHECKS PLAYTIME AND ADD TO A COUNTER - CAN I USE SOMETHING SIMILAR?

var counter = 0;
var currentIndex_inc = 0;
function onProgress() {

if(player.currentTime() <= 1){
    counter = 0;
}

//-- ------------------------------------- -->
// ----- COUNTER - If track plays longer than 30 seconds - add 1 --------
//-- ------------------------------------- -->
if(player.currentTime() >= 30  && trackURL != ''){
    if(counter==0){
    counter = 1;
    var playlist_name = "<?php echo $playlist ; ?>";
    var play_type = "<?php echo $type ; ?>";
    var trackURL = player.currentSrc();
    track_source = trackURL.src ;

        if(typeof(track_source)==="undefined"){
         track_source = trackURL;
        };

    $.ajax({
        type: "POST",
        url: "_inc/2018_process_counter.php",
        dataType: "text",
            data: { 
                playlist_name: playlist_name,track_source:track_source }
    }).done(function( data ) {
});
    }
    return false;
}

Logic:
If (video link does not start || video link == live){ 
   skip 
} else if (video link does start || video link == dead) {
   play 
}

使用Queru列出代码-基于成功的答复。答案仅适用于单个ID，而不适用于列表...请参见下面的代码:

if ($result_a->num_rows > 0) {
        // output data of each row
        while($row = $result_a->fetch_assoc()) {
            $id = $row['id'];
            $share_key = $row['share_key'];
            echo $row['id'];
            echo '<br>';
            echo $row['artist'];
            echo '<br>';
            echo $row['title'];
            echo '<br>';
            echo $row['source_url'];
            echo '<br>';            
            $my_link = $row['source_url'];

            $testlink = substr($my_link, strrpos($my_link, '/' )+1)."\n";

            echo '<p style="color:#ff0000">';
            echo $testlink;
            echo '</p>';            

            //# is ERROR = https://www.youtube.com/watch?v=R5mpcDWpYSA
            // $url = "https://www.youtube.com/oembed?url=https://www.youtube.com/watch?v=R5mpcDWpYSA"; //# test video deleted.

            //# is OK = https://www.youtube.com/watch?v=mLuh_O4mYbA
            $url = "https://www.youtube.com/oembed?url=https://www.youtube.com/watch?v=".$testlink; //# test working (not deleted).
            echo $url;
            echo '<br>';
            try
            {
                set_error_handler(function() { /* # temp ignore Warnings/Errors */ });

                $fop = fopen($url, "rb");
                if ( !$fop && $fop==false) { throw new Exception(); }

                restore_error_handler(); //# restore Warnings/Errors

                echo "OK 200 ::: Youtube video was found";
            }
            catch ( Exception $e ) 
            { echo "Error 404 ::: Youtube video not found (deleted or bad link)"; }




            echo '<hr>';









        }
    } else {
        // echo "0 results";
    }

Answer 1

"...YouTube will disable certain videos due to copyright or judgement issues, but the links are still in my list. Can anyone recommend a JS or other solution or article that sees if the video link does not start in x amount of time to initiate a skip or next action. Please advise."

由于您已经在使用PHP代码，因此以下步骤是一种可能的选择:

1)向 https://www.youtube.com/oembed? + Youtube video URL发出请求。

请求示例:

https://www.youtube.com/oembed?url=https://www.youtube.com/watch?v=R5mpcDWpYSA

2)使用 fopen检查视频可用性。请注意， file_exists($url)在Youtube服务器上无法正常使用(即使视频本身已被删除，它们也总是返回某些页面内容)。

以下是可测试的示例代码:
(将根据视频状态回显“ OK 200”或“ ERROR 404”)

<?php

    //# is ERROR = https://www.youtube.com/watch?v=R5mpcDWpYSA
    $url = "https://www.youtube.com/oembed?url=https://www.youtube.com/watch?v=R5mpcDWpYSA"; //# test video deleted.

    //# is OK = https://www.youtube.com/watch?v=mLuh_O4mYbA
    //$url = "https://www.youtube.com/oembed?url=https://www.youtube.com/watch?v=mLuh_O4mYbA"; //# test working (not deleted).

    try
    {
        set_error_handler(function() { /* # temp ignore Warnings/Errors */ });

        $fop = fopen($url, "rb");
        if ( !$fop && $fop==false) { throw new Exception(); }

        restore_error_handler(); //# restore Warnings/Errors

        echo "OK 200 ::: Youtube video was found";
    }
    catch ( Exception $e ) 
    { echo "Error 404 ::: Youtube video not found (deleted or bad link)"; }

?>

选项2:

您还可以通过将 file_get_contents与对Youtube的 get_video_info?的请求一起使用来获得相同的结果。

请求示例:

https://www.youtube.com/get_video_info?video_id=R5mpcDWpYSA

示例代码:

<?php

    //# ERROR = https://www.youtube.com/watch?v=R5mpcDWpYSA
    $url = "https://www.youtube.com/get_video_info?video_id=R5mpcDWpYSA"; //# test video deleted.

    //# OK = https://www.youtube.com/watch?v=mLuh_O4mYbA
    //$url = "https://www.youtube.com/get_video_info?video_id=mLuh_O4mYbA"; //# test working (not deleted).

    $src = file_get_contents($url);

    //# find text... playabilityStatus%22%3A%7B%22status%22%3A%22OK ...
    $str1 = "playabilityStatus%22%3A%7B%22status%22%3A%22";
    $pos = strpos($src, $str1);

    $result = substr( $src, $pos + (strlen($str1)), 5);

    if( $result{0} == "O" && $result{1} == "K" )
    { echo "OK 200 ::: Youtube video was found"; }
    else 
    { echo "Error 404 ::: Youtube video not found (deleted or bad link)"; }

?>