Question

我正在写一个网站（这是我第一次没有java ..）而且做得很简单。我调用了一个javascript函数“changeVideo（）”，它向php页面“GetAVideo.php”发出请求，该页面返回一个随机选择的视频的URL（在我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，但今天，当我加载我的页面时，视频未加载，因为GET请求失败： “NS_ERROR_FAILURE：失败 xhr_object.send（）;“

我无法理解为什么，这是我的来源：的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！

Answer 1

答案：我不得不使用相对网址而不是绝对。 Firefox认为它是一个跨站点脚本。

我改变了： var request =“http://ogdabou.com/php/GetAVideo.php”;

要 var request =“/php/GetAVideo.php”;

然后： VideoJS播放器说： “不支持内容类型http text / plain”

我通过将“AddType video / webm .webm”添加到根文件夹上的 .htaccess文件来修复此问题。

Javascript Http Get Request错误：NS_ERROR_FAILURE

1 个答案: