javascript - php和js重定向错误

Question

我正在尝试为 YouTube 播放列表做随机化器，我遇到了这个问题，当我“重定向”到我应该重定向的页面时，它实际上将我重定向到“www.youtube.com”而不是“www.youtube.com/watch?v={videoID}”就像我将重定向设置为“www.youtube.com/watch?v=”而不带视频 ID 一样。

有人可以帮助我获得我想要的重定向吗？

PS:代码如下

<?php 
    srand(12);
    $x=mt_rand(1,27);
        $y=$x-1;
        $index=strval($y);
    if($index==""){$index="0";}else{$index=strval($y);}
        $list="PLY3Z88WiyNeqKK2NDtx4ajmvntxmXKVG5";
        $v="0";
        switch($x){
            case 1:
                $v="tLkEFH3KFVI";
            break;

            case 2:
                $v="ie84bcGoSUI";
            break;
                case 3:
                $v="kLKTlyjS8pU";
            break;
            case 4:
                $v="-fwtEE0w6DM";
                break;
            case 5:
                $v="7MPIRR0Fs_g";
                break;
            case 6:
                $v="vojFyoClC1s";
                break;
            case 7:
                $v="0fUv2xNAwIE";
                break;
            case 8:
                $v="fkCxqd3OWD8";
                break;
            case 9:
                $v="5RCedaSoMHk";
                break;
            case 10:
                $v="ME1RKZXxWbU";
                break;
            case 11:
                $v="p7MFuczjjAE";
                break;
            case 12:
                $v="nz5ZI6iDDCc";
                break;
            case 13:
                $v="-HlvqQtmoRY";
                break;
            case 14:
                $v="TIGuXAV9OyA";
                break;
            case 15:
                $v="RQG3pG1prVg";
                break;
            case 16:
                $v="A6GUSRu5_Zs";
                break;
            case 17:
                $v="uL_B_Sd0Jyo";
                break;

            case 18:
                $v="eVNASb7mg8c";
                break;
            case 19:
                $v="9KbxBDKDL-U";
                break;
            case 20:
                $v="_uRmVNJuTVI";
                break;
            case 21:
                $v="Ba_iPeWWJLg";
                break;
            case 22:
                $v="P8zFLKnOBZ";
                break;
            case 23:
                $v="5QwmpQ1pw9E";
                break;
            case 24:
                $v="U54nqXz9icA&";
                break;
            case 25:
                $v="o5mV1rbZjs";
                break;
            case 26:
                $v="mntpfq2ajFQ";
                break;
            case 27:
                $v="aqvtL-LoSXA";
                break;
            default:
                $v="_uRmVNJuTVI";
        }

?>

    <center><img src="redir.png" alt="redirecting"/></center>



<script>
    var kk='<?php print( json_encode($v)); ?>';
    var zz="https://www.youtube.com/watch?v="+kk
 setTimeout(function(){document.location.href=zz},2000)
</script>

Answer 1

您的 JavaScript 中似乎存在语法错误。

缺少分号:

 var zz="https://www.youtube.com/watch?v="+kk; //here
 console.log(zz); //what does this show? 
 setTimeout(function(){document.location.href=zz},2000)