Guzzle向Gmail API发送HTTP请求，'Invalid resource type: object'

Question

我使用 Guzzle 向 Gmail API 发送 HTTP 请求。

它抛出未捕获的异常"无效参数异常"，并带有消息"无效的资源类型：对象"看来我给GuzzleHttpPsr7Request()班喂错了？

    require_once __DIR__.'/vendor/autoload.php';
    $client = new Google_Client();
    $client->setScopes("https://www.googleapis.com/auth/gmail.readonly");
    putenv('GOOGLE_APPLICATION_CREDENTIALS=sample.json');
    $client->useApplicationDefaultCredentials();
    // returns a Guzzle HTTP Client
    $httpClient = $client->authorize();
    $data = new stdClass; 
    $data-> topicName ='projects/sample.com:sample/topics/topic';   
    $data-> labelIds = ["INBOX"];
    $data-> labelFilterAction = 'include';
    $request = new GuzzleHttpPsr7Request('POST','https://www.googleapis.com/gmail/v1/users/[e-mail]/watch',['Content-type'=>'application/json'],$data);
    $response = $httpClient->send($request);
    var_dump($response);
//------------output----------------------
Uncaught exception 'InvalidArgumentException' with message 'Invalid resource type: object'

感谢安德鲁·诺兰和丹尼斯·索拉科维奇，现在$request似乎格式正确。

以下是更新：我正在使用服务帐户凭据来监视新邮件，有谁知道我哪里做错了？非常感谢。

require_once __DIR__.'/vendor/autoload.php';
$client = new Google_Client();
$client->setScopes("https://www.googleapis.com/auth/gmail.readonly");
putenv('GOOGLE_APPLICATION_CREDENTIALS=sample.json');
$client->useApplicationDefaultCredentials();
$gmailService = new Google_Service_Gmail($client);
// returns a Guzzle HTTP Client
$httpClient = $client->authorize();
$data = new stdClass; 
$data-> topicName ='projects/sample.com:sample/topics/topic';   
$data-> labelIds = ["INBOX"];
$data-> labelFilterAction = 'include';
$data = json_encode((array)$data);
$request = new GuzzleHttpPsr7Request('POST', 'https://www.googleapis.com/gmail/v1/users/[e-mail]/watch',["Content-type" =>"application/json"],$data);
$response = $httpClient->send($request);
echo ($response->getBody());
//--------------output---------------------
{ "error": { "errors": [ { "domain": "global", "reason": "failedPrecondition", "message": "Bad Request" } ], "code": 400, "message": "Bad Request" } }

Answer 1

好的，我现在明白了，我错过了这一步：

您需要向 授予发布权限

serviceAccount:[e-mail] 在云发布/订阅开发人员控制台权限界面

然后通过发送直接 HTTP 请求与 Guzzle 的代码：

$client = new Google_Client();
$client->setAuthConfig('xxxx-b2838442f7ad.json'); // downloaded json file.
$client->setScopes(array("https://www.googleapis.com/auth/gmail.readonly"));
$user_to_impersonate = '[e-mail]'; //very important
$client->setSubject($user_to_impersonate); //important
$httpClient = $client->authorize();
$data = new stdClass; 
$data-> topicName ='projects/sample.com:sample/topics/topic';   
$data-> labelIds = ["INBOX"];
$data-> labelFilterAction = 'include';
$data = json_encode($data);
$request = new GuzzleHttpPsr7Request('POST', 'https://www.googleapis.com/gmail/v1/users/[e-mail]/watch',["Content-type" =>"application/json"],$data);
$response = $httpClient->send($request);
echo($response->getBody());

// ------------------output-----------------------------------
    { "historyId": "31325", "expiration": "1496503166524" }

或者Gmail类的代码-这更难，您需要研究User.php并WatchRequest.php，如果不是这篇文章，像我这样的业余爱好者就没有机会：

require_once __DIR__.'/vendor/autoload.php';
$client = new Google_Client();
$client->setAuthConfig('xxxx-b2838442f7ad.json'); // downloaded json file.
$client->setScopes(array("https://www.googleapis.com/auth/gmail.readonly"));
$user_to_impersonate = '[e-mail]'; //very important
$client->setSubject($user_to_impersonate); //important
$watchreq = new Google_Service_Gmail_WatchRequest;
$watchreq->setLabelIds(array('INBOX'));
$watchreq->setTopicName('projects/sample.com:sample/topics/topic');
$watchreq->setLabelFilterAction('include');
$gmailService = new Google_Service_Gmail($client);
$response =$gmailService->users->watch('me', $watchreq);
echo($response->expiration)."<br/>";
echo($response->historyId);
//---------------------output--------------------------
1496505109513
31331