上传到YouTube时获取TOKEN_INVALID

时间:2012-10-01 13:18:55

标签: php youtube-api zend-gdata

我尝试使用Zend_Gdata(Zend Framework 1.12.0)通过API将视频上传到YouTube。我没有遇到直接上传工作的问题,但基于浏览器的上传总是给我400-INVALID TOKEN错误。我很确定我必须遗漏一些重要的东西,但要小到不能注意到它。

这涉及两个文件:

的index.php

<?php
$youTubeAPIKey = '<API_Key>';
$username = '<user>';
$password = '<pass>';

set_include_path(get_include_path().PATH_SEPARATOR.__DIR__."/vendor");
require_once 'Zend/Loader/Autoloader.php';
Zend_Loader_Autoloader::getInstance();

try
{
    $authenticationURL= 'https://www.google.com/accounts/ClientLogin';
    $httpClient = Zend_Gdata_ClientLogin::getHttpClient(
              $username,
              $password,
              $service = 'youtube',
              $client = null,
              $source = 'BrowserUploaderTest', // a short string identifying your application
              $loginToken = null,
              $loginCaptcha = null,
              $authenticationURL);

    $yt = new Zend_Gdata_YouTube($httpClient, "browser upload test", "Test version 0.1", $youTubeAPIKey);
    $videoEntry = new Zend_Gdata_YouTube_VideoEntry();

    $videoEntry->setVideoTitle("Test movie");
    $videoEntry->setVideoDescription("This is a test movie");
    $videoEntry->setVideoPrivate();

    // @todo This must be a valid YouTube category, how to get a list of valid categories?
    $videoEntry->setVideoCategory('Autos');
    $videoEntry->setVideoTags('cars, funny');

    // Get an upload token
    $tokenHandlerUrl = 'http://gdata.youtube.com/action/GetUploadToken';
    $tokenArray = $yt->getFormUploadToken($videoEntry, $tokenHandlerUrl);
    $token = $tokenArray['token'];
    $url = $tokenArray['url'];
    // print "Token value: {$tokenArray['token']}\n url: {$tokenArray['url']}\n";
    $nextUrl = "http://" . $_SERVER['HTTP_HOST'] . "/uploadDone.php";
}
catch (Zend_Gdata_App_HttpException $httpException)
{
    echo $httpException->getRawResponseBody();
}
catch (Zend_Gdata_App_Exception $e) {
    echo $e->getMessage();
}
catch (Exception $e)
{
    print $e->getTraceAsString();
}
?><!DOCTYPE html>
<html>
    <head>
        <title>Testing Youtube upload</title>
    </head>

    <body>
        <table>
            <tr>
                <td>
                    Url:
                </td>
                <td>
                    <?= $url ?>
                </td>
            </tr>
            <tr>
                <td>
                    Token:
                </td>
                <td>
                    <?= $token ?>
                </td>
            </tr>
        </table>
        <form action="<?= $url ?>.?nexturl=<?= urlencode($nextUrl) ?>" enctype="multipart/form-data" method="post">
            <input name="token" type="hidden" value="<?= $token ?>" />
            <input name="file" type="file" />
            <input type="submit" value="Upload file" />
        </form>
    </body>
</html>

uploadDone.php

<?php
print nl2br(print_r($_GET, true));
print nl2br(print_r($_POST, true));

我已经在Stack Overflow上进行了搜索,并花了几个小时在Google上搜索,但没有找到解决它的任何内容,这让我相信我错过了一些简单的东西。任何帮助将不胜感激。

备注: 此代码仅用于测试API使用情况,主要来自Google的开发人员指南(https://developers.google.com/youtube/2.0/developers_guide_php#Browser_based_Upload),并提供了一些帮助。 Yii框架文档(http://www.yiiframework.com/wiki/375/youtube-api-v2-0-browser-based-uploading/)。生产代码将以更有条理的方式重写,但目前并不重要。

1 个答案:

答案 0 :(得分:1)

您的action="<?= $url ?>.?nexturl=<?= urlencode($nextUrl) ?>"看起来很可疑;在您的.变量被评估之后,是否存在错误的$url字符,弄乱了网址?