GuzzleHttp 异常信息为什么会被截取(truncated...)？

解决方法

 try {
            $response = $this->httpClient->request('POST', $request_url, [
                'form_params' => $form,
                   
            ]);

            $body = $response->getBody()->getContents();
        } catch (ServerException $e) {
            $_error_code = $e->getCode();
            $_error_message = $e->getResponse()->getBody()->getContents();
        } catch (ClientException $e) {
            $_error_code = $e->getCode();
            $_error_message = $e->getResponse()->getBody()->getContents();
        } catch (RequestException $e) {
            $_error_code = $e->getCode();
            $_error_message = $e->getResponse()->getBody()->getContents();
        }