php - 基于YouTube api v3浏览器的上传引发400错误

Question

我正在基于youtube浏览器的视频上传中工作。但它抛出一个错误。
Fatal error: Uncaught exception 'Zend_Gdata_App_HttpException' with message 'Expected response code 200, got 400 <?xml version='1.0' encoding='UTF-8'?><errors><error><domain>yt:validation</domain><code>invalid_value</code><location type='xpath'>media:group/media:category[@scheme='http://gdata.youtube.com/schemas/2007/categories.cat']/text()</location></error></errors>' in /home/xxx/public_html/yyy.com/up/Zend/Gdata/App.php:714 Stack trace: #0 /home/xxx/public_html/yyy.com/up/Zend/Gdata.php(219): Zend_Gdata_App->performHttpRequest('POST', 'http://gdata.yo...', Array, '<atom:entry xml...', 'application/ato...', NULL) #1 /home/xxx/public_html/yyy.com/up/Zend/Gdata/App.php(905): Zend_Gdata->performHttpRequest('POST', 'http://gdata.yo...', Array, '<atom:entry xml...', 'application/ato...') #2 /home/xxx/public_html/yyy.com/up/Zend/Gdata/YouTube.php(704): Zend_Gdata_App->post(Object(Zend_Gdata_YouTube_VideoEntry), 'http://gdata.yo...') #3 /home/xxx/public_html/yyy.com/up/video_upload. in /home/xxx/public_html/yyy.com/up/Zend/Gdata/App.php on line 714
码:

<?php
// video_upload.php


require_once 'Zend/Loader.php'; // the Zend dir must be in your include_path
Zend_Loader::loadClass('Zend_Gdata_YouTube');
$yt = new Zend_Gdata_YouTube();

// Define the authentication that will be used
Zend_Loader::loadClass('Zend_Gdata_ClientLogin'); 

// Authenticate
$authenticationURL= 'https://www.google.com/accounts/ClientLogin';
$httpClient = 
Zend_Gdata_ClientLogin::getHttpClient(
          $username = "example@gmail.com",
          $password = "example",
          $service = 'youtube',
          $client = null,
          $source = 'HTML SOURCE CODE SNIPPET',
          $loginToken = null,
          $loginCaptcha = null,
          $authenticationURL);

$applicationId = 'YOUR APPLICATION ID';
$clientId = 'Upload videos to youtube using the youtube API';
$developerKey = 'A...........................0';
$yt = new Zend_Gdata_YouTube($httpClient, $applicationId, $clientId, $developerKey);



// create a new VideoEntry object
$myVideoEntry = new Zend_Gdata_YouTube_VideoEntry();

$myVideoEntry->setVideoTitle('My');
$myVideoEntry->setVideoDescription('video');
// The category must be a valid YouTube category!
$myVideoEntry->setVideoCategory('news');

// Set keywords. Please note that this must be a comma-separated string
// and that individual keywords cannot contain whitespace
$myVideoEntry->SetVideoTags('news');

$tokenHandlerUrl = 'http://gdata.youtube.com/action/GetUploadToken';
$tokenArray = $yt->getFormUploadToken($myVideoEntry, $tokenHandlerUrl);
$tokenValue = $tokenArray['token'];
$postUrl = $tokenArray['url'];


$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, $postUrl."?nexturl=//example.com/up");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLINFO_HEADER_OUT, true);
curl_setopt($ch, CURLOPT_POST, true);
// same as <input type="file" name="file">
$post = array("file"=>"@".$VideoFile['tmp_name'], "token"=>$tokenValue);
curl_setopt($ch, CURLOPT_POSTFIELDS, $post); 
$response = curl_exec($ch);
$info = curl_getinfo($ch);

echo $info;
?>

形成:

    <form action="video_upload.php" method="post" enctype="multipart/form-data">
    <input type="file" name="uploadedfile"><br>
    <input type="submit" value="Upload File to Server">
    </form>

我想直接将视频上传到我的youtube帐户。以便用户可以将视频上传到我的youtube帐户。

所以我没有在 $applicationId和 $clientId中添加任何内容。

我认为 $applicationId和 $clientId需要做一些事情。

谢谢

更新:

如 Wanny Miarelli 所建议
我将 news更改为 News，现在没有错误显示，但视频也没有上传。

尝试 print_r($info)并得到了这个

Array
(
   [url] => http://uploads.gdata.youtube.com/action/FormDataUpload/AIwbFAS1SkX04C09Re5sWeYXb562vc7W5KHcRu0uZcJ7v-A4nX_21zVwIOxdeBtp_s4tj-UcnZxJaN-Zd7nKVUfWwtWxW2YEPg?nexturl=//example.com/up
    [content_type] => 
    [http_code] => 0
    [header_size] => 0
    [request_size] => 0
    [filetime] => -1
    [ssl_verify_result] => 0
    [redirect_count] => 0
    [total_time] => 0.024856
    [namelookup_time] => 0.018251
    [connect_time] => 0.024751
    [pretransfer_time] => 0
    [size_upload] => 0
    [size_download] => 0
    [speed_download] => 0
    [speed_upload] => 0
    [download_content_length] => -1
    [upload_content_length] => -1
    [starttransfer_time] => 0
    [redirect_time] => 0
    [certinfo] => Array
        (
        )

    [redirect_url] => 
)

Answer 1

是的，似乎是类别问题。尝试将“新闻”更改为“新闻”

在此处查找完整的类别列表

https://developers.google.com/youtube/2.0/reference#Region_specific_feeds

这是直接下载到.cat文件

http://gdata.youtube.com/schemas/2007/categories.cat

这是Goolge在视频上传后(成功或失败)应发送给您的信息

After uploading the video file, the user is redirected to the nexturl specified in your form. If the upload was successful, YouTube appends id and status parameters to the URL as shown in the following example:

http://www.example.com/youtube_uploads?status=200&id=JPF-DXF7hzc如果
上传失败，YouTube会附加状态和代码
URL的参数，以帮助您诊断失败的原因。
引用指南提供了有关这些参数的更多信息。

好的，我建立了一个zend框架，这个代码对我有用

*此代码仅是解释性目的-不安全也不理想*

用户想要通过一个步骤表单上传视频，这是解决方案

    <?php

            //single file script, need to know if we are in a POST or GET request, this needs to be changed, dividing the script into a separate file 

            if(isset($_FILES['file']['name'])){


              Zend_Loader::loadClass('Zend_Gdata_YouTube');
              Zend_Loader::loadClass('Zend_Gdata_ClientLogin');

              $authenticationURL= 'https://www.google.com/accounts/ClientLogin';

              $httpClient = Zend_Gdata_ClientLogin::getHttpClient(
                            $username = '',
                            $password = '',
                            $service = 'youtube',
                            $client = null,
                            $source = 'TestApp', // a short string identifying your application
                            $loginToken = null,
                            $loginCaptcha = null,
                            $authenticationURL);


              $applicationId = 'TestAPP';
              $developerKey = "---";
              $clientId = "";

              // Note that this example creates an unversioned service object.
              // You do not need to specify a version number to upload content
              // since the upload behavior is the same for all API versions.
              $yt = new Zend_Gdata_YouTube($httpClient, $applicationId, $clientId, $developerKey);

              // create a new VideoEntry object
              $myVideoEntry = new Zend_Gdata_YouTube_VideoEntry();

              $myVideoEntry->setVideoTitle($_POST['Title']);
              $myVideoEntry->setVideoDescription($_POST['Description']);
              // The category must be a valid YouTube category!
              $myVideoEntry->setVideoCategory('News');

              // Set keywords. Please note that this must be a comma-separated string
              // and that individual keywords cannot contain whitespace
              $myVideoEntry->SetVideoTags('News');

              $tokenHandlerUrl = 'http://gdata.youtube.com/action/GetUploadToken';
              $tokenArray = $yt->getFormUploadToken($myVideoEntry, $tokenHandlerUrl);
              $tokenValue = $tokenArray['token'];
              $postUrl = $tokenArray['url'];

              // place to redirect user after upload
              $nextUrl = 'http://localweb/';

              $target_url = 'http://127.0.0.1/accept.php';
            //This needs to be the full path to the file you want to send.
                $file_name_with_full_path = realpath('C:\video.flv');
                      /* curl will accept an array here too.
                       * Many examples I found showed a url-encoded string instead.
                       * Take note that the 'key' in the array will be the key that                    shows up in the
                       * $_FILES array of the accept script. and the at sign '@' is required before the
                       * file name.
                       */

                 // Post the video to Youtube Server
                $post = array('token' => $tokenValue,'file_contents'=>'@'. $_FILES["file"]["tmp_name"]);

                $ch = curl_init();
                curl_setopt($ch, CURLOPT_URL, $postUrl . '?nexturl=' . $nextUrl);
                curl_setopt($ch, CURLOPT_POST,1);
                curl_setopt($ch, CURLOPT_POSTFIELDS, $post);
                $result=curl_exec ($ch);
                curl_close ($ch);
                echo $result;


    }else{

      $self = $_SERVER['PHP_SELF'];

      $form = '<form action="' . $self  . '"'.
            ' method="post" enctype="multipart/form-data">'. 
            '<input name="file" id="file" type="file"/>'. 
            '<input name="token" id="token" type="hidden" value=""/>'.
            '<input name="Description" id="Description" />'.
            '<input name="Title" id="Title" />'.
            '<input value="Upload Video File" type="submit" />'. 
            '</form>';

            echo $form;
    }