Question

我正在尝试使用PHP将视频上传到Youtube。我使用的是Youtube API v3，我使用的是最新检出的Google API PHP客户端库源代码我正在使用上面给出的示例代码 https://code.google.com/p/google-api-php-client/执行身份验证。身份验证很顺利但是当我尝试上传视频时，我得到Google_ServiceException，错误代码为500，消息为null。

我看过前面提到的以下问题： Upload video to youtube using php client library v3但接受的答案并未说明如何指定要上传的文件数据我发现了另一个类似的问题Uploading file with Youtube API v3 and PHP，其中在评论中提到categoryId是强制性的，因此我尝试在片段中设置categoryId，但它仍然给出相同的异常。

我还提到了文档站点上的Python代码（https://developers.google.com/youtube/v3/docs/videos/insert），但我无法在客户端库中找到函数next_chunk。但是我尝试将一个循环（在代码片段中提到）重试以获取错误代码500，但是在所有10次迭代中我都得到了同样的错误。

以下是我正在尝试的代码段：

$youTubeService = new Google_YoutubeService($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>";
}

我做错了吗？

Answer 1

我能够使用以下代码完成上传工作：

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();
    }
}

Answer 2

我意识到这是旧的，但这是文档中的答案：

    // 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);

Answer 3

我也意识到这是旧的，但是因为我从GitHub克隆了最新版本的php-client，我遇到了Google_Service_YouTube_Videos_Resource::insert() - 方法的问题。

我会传递一个数组，"data" => file_get_contents($pathToVideo)和"mimeType" => "video/mp4"设置为insert()方法的参数，但我仍然继续（400）BadRequest 作为回报。

通过我在\Google\Service\Resource.php中找到的Google代码进行调试和阅读，对数组密钥"uploadType"进行了检查（在第179-180行），这将启动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);

如果我没记错的话，使用PHP-api的0.6版本就不需要uploadType参数。这可能仅适用于直接上传样式，而不适用于任何一天的答案中显示的可恢复上传。

Answer 4

答案是通过Google PHP client libraries使用Google_Http_MediaFileUpload。

以下是示例代码：https://github.com/youtube/api-samples/blob/master/php/resumable_upload.php

使用Youtube API V3和PHP将视频上传到Youtube

4 个答案: