我想上传使用PHP的Youtube视频。 我使用的Youtube API v3和我使用谷歌API PHP客户端库的最新签出的源代码。
我使用给出的示例代码
https://code.google.com/p/google-api-php-client/进行认证。 该认证经过精细，但当我尝试上传的视频，我得到Google_ServiceException ，错误代码500和消息为空。
我看了一下以下问题刚才问： 上传视频的YouTube使用PHP客户端库V3到 ，但接受的答案没有描述如何指定文件数据上传。
我发现了另一个类似的问题与YouTube API v3和PHP上传文件 ，凡在评论中提到，的categoryId是强制性的，因此我试图在段设置的categoryId但仍给出了同样的异常。
我还提到了Python代码的文档网站（ https://developers.google.com/youtube/v3/docs/videos/insert ），但我找不到客户端库功能next_chunk。 但我试图把一个循环（代码段提到的）重试对收到错误代码500，但在所有的10次迭代，我得到了同样的错误。
以下是代码片段我想：
$youTubeService = new Google_YoutubeService($client);
if ($client->getAccessToken()) {
print "Successfully authenticated";
$snippet = new Google_VideoSnippet();
$snippet->setTitle = "My Demo title";
$snippet->setDescription = "My Demo descrition";
$snippet->setTags = array("tag1","tag2");
$snippet->setCategoryId(23); // this was added later after refering to another question on stackoverflow
$status = new Google_VideoStatus();
$status->privacyStatus = "private";
$video = new Google_Video();
$video->setSnippet($snippet);
$video->setStatus($status);
$data = file_get_contents("video.mp4"); // This file is present in the same directory as the code
$mediaUpload = new Google_MediaFileUpload("video/mp4",$data);
$error = true;
$i = 0;
// I added this loop because on the sample python code on the documentation page
// mentions we should retry if we get error codes 500,502,503,504
$retryErrorCodes = array(500, 502, 503, 504);
while($i < 10 && $error) {
try{
$ret = $youTubeService->videos->insert("status,snippet",
$video,
array("data" => $data));
// tried the following as well, but even this returns error code 500,
// $ret = $youTubeService->videos->insert("status,snippet",
// $video,
// array("mediaUpload" => $mediaUpload);
$error = false;
} catch(Google_ServiceException $e) {
print "Caught Google service Exception ".$e->getCode()
. " message is ".$e->getMessage();
if(!in_array($e->getCode(), $retryErrorCodes)){
break;
}
$i++;
}
}
print "Return value is ".print_r($ret,true);
// We're not done yet. Remember to update the cached access token.
// Remember to replace $_SESSION with a real database or memcached.
$_SESSION['token'] = $client->getAccessToken();
} else {
$authUrl = $client->createAuthUrl();
print "<a href='$authUrl'>Connect Me!</a>";
}
难道我做错了什么吗？
我能得到上传使用下面的代码工作：
if($client->getAccessToken()) {
$snippet = new Google_VideoSnippet();
$snippet->setTitle("Test title");
$snippet->setDescription("Test descrition");
$snippet->setTags(array("tag1","tag2"));
$snippet->setCategoryId("22");
$status = new Google_VideoStatus();
$status->privacyStatus = "private";
$video = new Google_Video();
$video->setSnippet($snippet);
$video->setStatus($status);
$error = true;
$i = 0;
try {
$obj = $youTubeService->videos->insert("status,snippet", $video,
array("data"=>file_get_contents("video.mp4"),
"mimeType" => "video/mp4"));
} catch(Google_ServiceException $e) {
print "Caught Google service Exception ".$e->getCode(). " message is ".$e->getMessage(). " <br>";
print "Stack trace is ".$e->getTraceAsString();
}
}
我意识到这是旧的，但这里的答案关闭文档：
// REPLACE this value with the path to the file you are uploading.
$videoPath = "/path/to/file.mp4";
$snippet = new Google_Service_YouTube_VideoSnippet();
$snippet->setTitle("Test title");
$snippet->setDescription("Test description");
$snippet->setTags(array("tag1", "tag2"));
// Numeric video category. See
// https://developers.google.com/youtube/v3/docs/videoCategories/list
$snippet->setCategoryId("22");
// Set the video's status to "public". Valid statuses are "public",
// "private" and "unlisted".
$status = new Google_Service_YouTube_VideoStatus();
$status->privacyStatus = "public";
// Associate the snippet and status objects with a new video resource.
$video = new Google_Service_YouTube_Video();
$video->setSnippet($snippet);
$video->setStatus($status);
// Specify the size of each chunk of data, in bytes. Set a higher value for
// reliable connection as fewer chunks lead to faster uploads. Set a lower
// value for better recovery on less reliable connections.
$chunkSizeBytes = 1 * 1024 * 1024;
// Setting the defer flag to true tells the client to return a request which can be called
// with ->execute(); instead of making the API call immediately.
$client->setDefer(true);
// Create a request for the API's videos.insert method to create and upload the video.
$insertRequest = $youtube->videos->insert("status,snippet", $video);
// Create a MediaFileUpload object for resumable uploads.
$media = new Google_Http_MediaFileUpload(
$client,
$insertRequest,
'video/*',
null,
true,
$chunkSizeBytes
);
$media->setFileSize(filesize($videoPath));
// Read the media file and upload it chunk by chunk.
$status = false;
$handle = fopen($videoPath, "rb");
while (!$status && !feof($handle)) {
$chunk = fread($handle, $chunkSizeBytes);
$status = $media->nextChunk($chunk);
}
fclose($handle);
// If you want to make other calls after the file upload, set setDefer back to false
$client->setDefer(false);
我也知道这是老了， 但我克隆了最新版的PHP客户端的从GitHub我跑与麻烦Google_Service_YouTube_Videos_Resource::insert() -方法。
我将通过以与阵列"data" => file_get_contents($pathToVideo)和"mimeType" => "video/mp4"设置为用于一个参数insert() -方法，但我仍然一直得到（400）错误请求回报。
调试并通过谷歌的代码阅读中，我发现在\Google\Service\Resource.php有一个检查（线179-180）对一个数组的键"uploadType" ，将启动Google_Http_MediaFielUpload对象。
$part = 'status,snippet';
$optParams = array(
"data" => file_get_contents($filename),
"uploadType" => "media", // This was needed in my case
"mimeType" => "video/mp4",
);
$response = $youtube->videos->insert($part, $video, $optParams);
如果我没有记错，与PHP-API 0.6版本不需要uploadType说法。 这可能仅适用于任何一天的回答显示了直接上传的风格，而不是继续上传。
答案是通过使用Google_Http_MediaFileUpload 谷歌PHP客户端库 。
这里的示例代码： https://github.com/youtube/api-samples/blob/master/php/resumable_upload.php
