php中隐藏url错误

我试图隐藏视频（videos/mymovie.mp4）URL，以使视频下载变得困难。我正在使用会话id和视频路径来获得加密的令牌。下面是我为卓越播放器（video.php）编写的代码。

<?php
     include('movie_url.php');
     echo '<video id="'.$movieId.'" class="sublime" poster="'.$cover_image_path.'" width="935" height="526" title="'.$movie_title.'" data-autoresize="fit" preload="none">
           <source src="moviesessionurl.php?video='.$hash.'" />
     </video>';
?> 

movie_url.php包含

<?php
    //Connect to database
    include("config.php");
    session_start();
    $sid = session_id();
    $movieId = $_SESSION['movie_id'];
    $sql = "SELECT * FROM `movies` WHERE intMovieID = ".$movieId;
    $result = mysqli_query($con,$sql);
    $row = mysqli_fetch_row($result);
    $movie_title = $row[1];
    $movie_desc = $row[7];
    $trailer_path = $row[8];
    $movie_path = $row[9];
    $cover_image_path = $row[10];
    $still_image_path = $row[11];
    $poster_image_path = $row[12];
    $movie_duration = $row[16];
    $hash = md5($movie_path.$sid);
    $_SESSION[$hash] = $movie_path;
    mysqli_close($con);                                
?>

moviesessionurl.php包含

<?php
    /* start the session */
    session_start();
    $path = $_SESSION[$_GET['video']];
    echo $path;
?>

当我直接在浏览器中运行moviesessionurl.php时，我会得到正确的视频url。但当我运行video.php时，它包含出色的播放器，视频不会播放！给出错误"Cant play video"

请帮我完成这件事。