php - 从 guzzle 捕获异常

Question

我正在使用 laravel，并且我已经设置了抽象类方法来从我调用的各种 API 中获取响应。但如果 API url 无法访问，则会抛出异常。我知道我错过了一些东西。任何帮助对我来说都是很大的。

$offers = [];
    try {
      $appUrl = parse_url($this->apiUrl);

      // Call Api using Guzzle
      $client = new Client('' . $appUrl['scheme'] . '://' . $appUrl['host'] . '' . $appUrl['path']);

      if ($appUrl['scheme'] == 'https') //If https then disable ssl certificate
        $client->setDefaultOption('verify', false);

      $request = $client->get('?' . $appUrl['query']);
      $response = $request->send();
      if ($response->getStatusCode() == 200) {
        $offers = json_decode($response->getBody(), true);
      }
    } catch (ClientErrorResponseException $e) {
      Log::info("Client error :" . $e->getResponse()->getBody(true));
    } catch (ServerErrorResponseException $e) {
      Log::info("Server error" . $e->getResponse()->getBody(true));
    } catch (BadResponseException $e) {
      Log::info("BadResponse error" . $e->getResponse()->getBody(true));
    } catch (\Exception $e) {
      Log::info("Err" . $e->getMessage());
    }

    return $offers;

Answer 1

您应该使用选项 'http_errors' => false, 设置 guzzehttp 客户端示例代码应该是这样的，document:guzzlehttp client http-error option explain

Set to false to disable throwing exceptions on an HTTP protocol errors (i.e., 4xx and 5xx responses). Exceptions are thrown by default when HTTP protocol errors are encountered.

$client->request('GET', '/status/500');
// Throws a GuzzleHttp\Exception\ServerException

$res = $client->request('GET', '/status/500', ['http_errors' => false]);
echo $res->getStatusCode();
// 500





$this->client = new Client([
    'cookies' => true,
    'headers' => $header_params,
    'base_uri' => $this->base_url,
    'http_errors' => false
]);

$response = $this->client->request('GET', '/');
if ($code = $response->getStatusCode() == 200) {
   try {
       // do danger dom things in here
   }catch (/Exception $e){
      //catch
   }

}