Javascript Http Get Request error : NS_ERROR_FAILURE


Javascript Http Get Request error : NS_ERROR_FAILURE

我正在写一个网站(这是我第一次没有java),所做的很简单。我调用一个javascript函数"changeVideo()",它向php页面"GetAVideo.php"发出请求,该页面将URL返回到随机选择的视频(在我的服务器上的视频文件之间选择)。

昨晚,我可以毫无问题地观看视频,但是今天,当我加载页面时,视频没有加载,因为GET请求失败:"NS_ERROR_FAILURE:失败xhr_object.send();"

我不明白为什么,这是我的消息来源:JavaScript:

function changeVideo() 
{
    console.log("changeVideo path:", path);
    var xhr_object = null;
    if (window.XMLHttpRequest) // Firefox 
        xhr_object = new XMLHttpRequest();
    else if (window.ActiveXObject) // Internet Explorer 
        xhr_object = new ActiveXObject("Microsoft.XMLHTTP");
    var request = "http://ogdabou.com/php/GetAVideo.php";
    console.log("request: ", request);
    xhr_object.open("GET", request, false);
    xhr_object.send();
    console.log("response: ", xhr_object.responseText);
    videoPlayer.src(xhr_object.responseText);
    videoPlayer.currentTime(0);
    videoPlayer.play();
    document.getElementById("videoTitle").innerHTML = xhr_object.responseText;
    return false;
};

.php:

<?php
    $dirname = '../videos';
    $videoList = array();
    $dir = opendir($dirname);
    if (count($_GET) > 0)
    {
        $folders=explode(";", $_GET['folders']);
        foreach ($folders as $videoFolder) {
            $fullPath = $dirname."/".$videoFolder;
            echo "Visiting $fullpath";
            $videoList = fillVideoList($fullPath, $videoList);
        }
    }
    else
    {
        $videoList = fillVideoList($dirname, $videoList);
    }
    //use join to get the paths.
    closedir($dir);
    $choosenOne = $videoList[rand(0, count($videoList) - 1)];
    $choosenOne = str_replace("../videos/","", $choosenOne);
    echo "http://ogdabou.com/videos/".$choosenOne;
?>
<?php
    // Fill the videoList with the given folder. Also visit subdirectories.
    function fillVideoList($folder, $videoList)
    {
        $path = $folder;
        $folder = opendir($folder);
        while($file = readdir($folder)) { 
                if($file != '.' && $file != '..')
                {
                        $fullPath = "$path/$file";
                        if (is_dir($fullPath))
                        {
                                $videoList = fillVideoList($fullPath, $videoList);
                        }
                        else if(pathinfo($fullPath, PATHINFO_EXTENSION) == "webm")
                        {
                                $videoList[] = $fullPath;
                        }
                }    
        }
        return $videoList;  
    }
?>

谢谢!

答案 :我不得不使用相对网址而不是绝对网址。Firefox 将其识别为跨站点脚本。

我改变了: var 请求 ="http://ogdabou.com/php/GetAVideo.php";

自 var request = "/php/GetAVideo.php";


然后:VideoJS播放器说:"不支持内容类型 HTTP 文本/纯文本"

我通过将"AddType video/webm .webm"添加到根文件夹上的 .htaccess 文件中来解决此问题。