Question

我想知道检查guzzle响应是否正常并避免阻止应用程序的最佳做法是什么。我使用PHP / Symfony，每次拨打电话时我都会这样做：

try {

    $response = $this->getClient()->request('GET', '/api/rest/contact/' . $email);

} catch (\Exception $e) {

    $logger = $this->get('monolog.logger.myapp');
    $logger->critical('New exception caught while getting user: ' . $e);
    throw new HttpException(406, "Error while getting user.");
}

if(isset($response) && $response->getStatusCode() == 200) {
    return $response->getBody()->getContents();
}

// if it's not 200 or the response is not set, I send a JsonResponse or a flash message to be used in a form for instance:

$this->addFlash('error', $this->get('translator')->trans('form.subscribe.fail', array(), 'messages'));

// or

return new JsonResponse(array('messages' => [0 => $this->get('translator')->trans('form.subscribe.fail', array(), 'messages')]), 400);

编辑以适应收到的答案：

try {

    $response = $this->getClient()->request('GET', '/api/rest/contact/' . $email);

} catch (\Exception $e) {

    $logger = $this->get('monolog.logger.myapp');
    $logger->critical('New exception caught while getting user: ' . $e);

    // the response is not 200 so I send a JsonResponse or a flash message to be used in a form for instance:

    $this->addFlash('error', $this->get('translator')->trans('form.subscribe.fail', array(), 'messages'));

    // or

    return new JsonResponse(array('messages' => [0 => $this->get('translator')->trans('form.subscribe.fail', array(), 'messages')]), 400);
}

return $response->getBody()->getContents();

Answer 1

＆＃34; OK＆＃34;取决于你的终点＆＃39;供应商。即使状态代码= 200（某些crapy API执行此操作），某些提供程序也可以响应错误。

基本上，如果状态代码发出有关错误的信号（状态代码＆gt; = 400），Guzzle默认会抛出异常。所以你不需要做额外的检查，只处理异常。

BTW，请查看this answer了解更多信息。

验证guzzle响应的最佳做法

1 个答案: