我正在开发基于youtube浏览器的视频上传。但它抛出了一个错误。
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
代码：
<?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 $info;
?>
形式：
<form action="video_upload.php" method="post" enctype="multipart/form-data">
<input type="file" name="uploadedfile"><br>
<input type="submit" value="Upload File to Server">
</form>
我想直接将视频上传到我的YouTube帐户。用户可以将视频上传到我的YouTube帐户..
所以我没有向$applicationId和$clientId添加任何内容。
我认为需要使用$applicationId和$clientId完成某些事情。
由于
更新
正如 Wanny Miarelli 建议的那样
我将news更改为News，现在没有显示错误，但视频也未上传。
尝试print_r($info)并获得此
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] =>
)
答案 0 :(得分：1)
是的，似乎是一个类别问题。尝试将“新闻”更改为“新闻”
查看完整的类别列表
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（成功或失败）后应该发送给您的
上传视频文件后，用户将被重定向到表单中指定的nexturl。如果上传成功，则为YouTube 将id和status参数附加到URL，如下所示 例如：
http://www.example.com/youtube_uploads?status=200&id=JPF-DXF7hzc如果 上传不成功，YouTube会附加状态和代码 URL的参数可以帮助您诊断失败的原因。 参考指南提供了有关这些参数的更多信息。
好的，我设置了一个zend框架，这段代码适用于我
* 此代码仅供参考 - 不安全且不优惠*
用户想要使用One Step Form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;
}