我是如何从guzzle6异常中得到身体的

Question

所以我有下面的代码

    try {
        $response = $client->request('POST', 'http://dev.api.example.com/v1/partners', [
            // params
        ]);
    } catch (ClientException $ex) {
        Debug::dump($ex);
        die;
    }

现在，我故意发送错误的数据进行测试，api发送了400段代码，因为缺少了一些东西，而且我正在使用try和catch块来捕获它。现在，我想向主体展示api返回的内容。我试过以下几种方法

$ex->getResponse()->getBody()

但它只返回以下内容。

GuzzleHttp\Psr7\Stream Object
(
    [stream:GuzzleHttp\Psr7\Stream:private] => Resource id #73
    [size:GuzzleHttp\Psr7\Stream:private] => 
    [seekable:GuzzleHttp\Psr7\Stream:private] => 1
    [readable:GuzzleHttp\Psr7\Stream:private] => 1
    [writable:GuzzleHttp\Psr7\Stream:private] => 1
    [uri:GuzzleHttp\Psr7\Stream:private] => php://temp
    [customMetadata:GuzzleHttp\Psr7\Stream:private] => Array
        (
        )

)

当api在postment中发送时

{
  "success": false,
  "error": {
    "code": 400,
    "message": "The name has already been taken.<br />The email field is required."
  }
}

Answer 1

好吧我已经弄清楚了。您所需要做的就是在例外情况下执行以下操作。

$ex->getResponse()->getBody()->getContents()

截至

http://docs.guzzlephp.org/en/latest/psr7.html