laravel GuzzleHttp Client 无法获取返回的错误信息

Client发送一些请求，当返回状态不是200的时候，无法获取完整错误信息

$client = new Client(['base_uri' => $this->getUri().'/order/aaaaaa','timeout' => 30,'verify' => false]);try {$response = $client->request('POST', '', ['headers' => ['Lang'=>'zh-CN','Authorization' => 'Bearer ' . $accessToken],'json' =>['channelCode' => '','exportId' => '',]]);$response = $response->getBody()->getContents();return $this->success('登录成功', collect([]));} catch (GuzzleException $e) {dd($e->getResponse()->getBody()->getContents());return $this->error($e->getMessage(), collect([]));}

最后当捕获到异常的时候，会出现下面这种现象

Client error: `POST http://192.168.11.1/api/order/aqedd` resulted in a `422 Unprocessable Content` response:{"success":false,"code":422,"locale":"zh-CN","message":"给定的数据不能为空","data":{"channelCode":["channel cod (truncated...)

可以看到，这并不是完整的报错信息，因为在最关键的代码排查处，出现了截取，截取的关键字就是truncated...，那么怎么获取到完整的信息呢
这样做

dd($e->getResponse()->getBody()->getContents());

使用上面的代码就可以获取到完整报错信息

被截断的原始是Guzzle代码里做了限制，来看Exception源码，源码来自文件

//.\vendor\guzzlehttp\guzzle\src\Exception\RequestException.php
public static function getResponseBodySummary(ResponseInterface $response){$body = $response->getBody();if (!$body->isSeekable()) {return null;}$size = $body->getSize();if ($size === 0) {return null;}$summary = $body->read(120);$body->rewind();if ($size > 120) {$summary .= ' (truncated...)';}// Matches any printable character, including unicode characters:// letters, marks, numbers, punctuation, spacing, and separators.if (preg_match('/[^\pL\pM\pN\pP\pS\pZ\n\r\t]/', $summary)) {return null;}return $summary;}

所以当$size超过了120个字符之后，就会用(truncated...)截取，当然我们不希望去改动GuzzleHttp的核心代码，所以还是使用上面的方法来获取完整的异常信息吧。