javascript - How to realize that when the user opens the video address, it does not play but downloads

<code><?php
    $file = "/tmp/视频.mp4";
    $filename = basename($file);
    header("Content-type: application/octet-stream");
    //处理中文文件名
    $ua = $_SERVER["HTTP_USER_AGENT"];
    $encoded_filename = rawurlencode($filename);
    if (preg_match("/MSIE/", $ua)) {
     header('Content-Disposition: attachment; filename="' . $encoded_filename . '"');
    } else if (preg_match("/Firefox/", $ua)) {
     header("Content-Disposition: attachment; filename*=\"utf8''" . $filename . '"');
    } else {
     header('Content-Disposition: attachment; filename="' . $filename . '"');
    }
    header("Content-Length: ". filesize($file));
    readfile($file);</code>