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如果视频“不可用”,是否有跳过播放列表中YouTube视频的修复程序

playlist youtube video

0

NetTemple · 技术社区 · 5 年前

没有解决方案,因为大量的谷歌搜索没有任何建议。

我的基本逻辑是,如果x秒后视频不播放,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 
}

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";
    }

0 回复 | 直到 5 年前

1

VC.One 5 年前

“…由于版权或判断问题,YouTube将禁用某些视频,但链接仍在我的列表中。有没有人可以推荐一个JS或者其他的解决方案或者文章,看看视频链接是否在x个时间段内没有启动来启动一个跳过或者下一个动作。请告知。”

既然已经涉及到PHP代码,那么一个可能的选择就是以下步骤:

https://www.youtube.com/oembed? + Youtube video URL .

请求示例:

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

fopen 检查视频可用性。附注a file_exists($url) 在Youtube服务器上无法正常工作(它们总是返回一些页面内容,即使视频本身已被删除)。

(将回音) OK 200 “或” ERROR 404 “,取决于视频状态…)

<?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)"; }

?>

方案2:

file_get_contents 向Youtube的 get_video_info? .

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)"; }

?>