Php 狂饮到莱佛api

大家好，我一直在为这个问题发愁。我正在从ryver访问一个api，以发送聊天后消息以进行自动通知。我正在关注这个文档，我正在使用Laravel5.1和Guzzle，这是我的代码，如果它有帮助的话

$client = new Client();
        $postData = \GuzzleHttp\json_encode(['JSON Payload' => ['body' => 'test123']]);



        $options = [
            'json' => $postData
        ];

        $request = $client->post('https://somecompany.ryver.com/api/1/odata.svc/workrooms(1099207)/Chat.PostMessage()', $options);
        $request->setHeader('Content-Type', 'application/json');
        $request->setHeader('Accept', 'application/json');
        $request->setHeader('Authorization', 'Basic Base64codehere');

        $response = $request->send();

---

它总是返回一个[status code]400，请帮助：（谢谢您，祝您度过愉快的一天！

您可以尝试使用此参数发送API请求

$postData = \GuzzleHttp\json_encode(['body' => 'test123']);

$options = [
    'JSON Payload' => $postData
];

修复了：）我只需要对JSON进行字符串化以使其正常工作，并将主体设置为JSON。这是密码

$client = new Client();
        $postData = '{
        "body":"**Update!**\n> ** Test success for ryver integration.",
          "extras": {
            "from": {
              "__descriptor":"Developer",
              "avatarUrl":"https://cdn2.f-cdn.com/ppic/4973381/logo/4389970/developer_avatar.png"
            } 
          }
        }';
        $request = $client->post('https://company.ryver.com/api/1/odata.svc/workrooms(1098712)/Chat.PostMessage()',[
            'Content-Type' => 'application/json',
            'Accept' => 'application/json',
            'Authorization' => 'Basic base64'
        ]);

        $request->setBody($postData);

        $response = $request->send();

---

响应主体有什么？我不能向@JonStirling发送任何响应，因为它抛出了一个异常。客户端错误响应[状态代码]400[原因短语]错误请求