Guzzle的客户端不会返回错误消息。只有＆＃34;结果＆＃34;退回

Question

我不知道为什么Guzzle的客户只返回＆＃34;结果＆＃34;值。如何获得完整的响应数据。

（1）method_A返回结果。 Guzzle的客户端返回结果。这没问题。

// HTTP request with curl

curl -X POST --data '{"jsonrpc":"2.0","method":"method_A","params":["xxx"],"id":1}' https://example.com/rpc
{"jsonrpc":"2.0","id":1,"result":"0x5209"}

// HTTP request with Guzzle

$client = new Client(['base_uri' => 'https://example.com/rpc']);
$response = $client->request('POST', '', ['json' => $params]);
echo json_encode($response) // 0x5209

（2）method_B返回错误。 Guzzle的客户端返回空而不是错误。这是我的问题。

// HTTP request with curl

curl -X POST --data '{"jsonrpc":"2.0","method":"method_B","params":["xxx"],"id":1}' https://example.com/rpc
{"jsonrpc":"2.0","id":1,"error":{"code":-32000,"message":"nonce too low"}}

// HTTP request with Guzzle

$client = new Client(['base_uri' => 'https://example.com/rpc']);
$response = $client->request('POST', '', ['json' => $params]);
echo json_encode($response) // empty

FYI。 https://example.com/rpc返回总是200个响应，即使响应数据由于某些原因而出错。

更新1

我尝试了以下内容。

Log::info($response);
Object of class GuzzleHttp\Psr7\Response could not be converted to string

Log::info(json_decode($response));
json_decode() expects parameter 1 to be string, object given

Log::info(var_dump($response));
<pre class='xdebug-var-dump' dir='ltr'>
<b>object</b>(<i>GuzzleHttp\Psr7\Response</i>)[<i>374</i>]
  <i>private</i> 'reasonPhrase' <font color='#888a85'>=&gt;</font> <small>string</small> <font color='#cc0000'>'OK'</font> <i>(length=2)</i>

...