How can I make Symfony2 ignore Guzzle Client bad response exception in my custom controller?

Question

   function order_confirmationAction($order,$token) { 

        $client = new \GuzzleHttp\Client();
        $answer  = $client->post("http://www.fullcommerce.com/rest/public/Qtyresponse",
                    array('body' => $order)
        );

        $answer  = json_decode($answer); 

        if ($answer->status=="ACK") {
            return $this->render('AcmeDapiBundle:Orders:ack.html.twig', array(
            'message'   => $answer->message,
        ));
        } else throw new \Symfony\Component\HttpKernel\Exception\HttpException(500, $answer->message);
}

If $client->post() response status code is an "Error 500" Symfony stops the script execution and throw new exception before the json decoding. How can I force Symfony to ignore $client->post() bad response and execute till the last if statement?

I believe that you will find your answer in this other question: http://stackoverflow.com/questions/17658283/catching-exceptions-from-guzzle — Eric Redon, Jul 30 '14 at 19:49
getEventDispatcher() and addListener() are deprecated function in the current Guzzle release! — EmaOnTheBlock, Jul 30 '14 at 20:15
The try/catch way of catching an error is still current, though :) — Eric Redon, Jul 30 '14 at 20:28

score 3 · Answer 1 · answered Jul 31 '14 at 10:35

            $client = new \GuzzleHttp\Client();
            try {
                $answer  = $client->post("http://www.fullcommerce.com/rest/public/Qtyresponse",
                        array('body' => $serialized_order)
                );
            }
            catch (\GuzzleHttp\Exception\ServerException $e) {

                if ($e->hasResponse()) {
                    $m = $e->getResponse()->json();
                    throw new \Symfony\Component\HttpKernel\Exception\HttpException(500, $m['result']['message']);
                }

            }

I solved like this. In that way I can access to responses of remote server even if it returns an error 500 code.

score 0 · Answer 2 · answered Jul 30 '14 at 21:03

Per Guzzle documentation:

Guzzle throws exceptions for errors that occur during a transfer.

Specifically, if the API responds with a 500 HTTP error, you shouldn't expect its content to be JSON, and you don't want to parse it, so you're better off re-throwing an exception from there already (or informing the user that something went wrong). I would suggest trying this out:

function order_confirmationAction($order, $token) { 
    $client = new \GuzzleHttp\Client();
    try {
        $answer  = $client->post("http://www.fullcommerce.com/rest/public/Qtyresponse",
            array('body' => $order)
        );
    }
    catch (Exception $e) {
        throw new \Symfony\Component\HttpKernel\Exception\HttpException(500, $e->getMessage());
    }

    $answer  = json_decode($answer); 

    if ($answer->status=="ACK") {
        return $this->render('AcmeDapiBundle:Orders:ack.html.twig', array(
            'message'   => $answer->message,
        ));
    } else {
        throw new \Symfony\Component\HttpKernel\Exception\HttpException(500, $answer->message);
    }
}

It is probably also a good idea to check for errors when JSON-decoding the response, because there could be surprises in the content you're getting (eg. wrong format, missing or unexpected fields or values, etc.).

I think you are right when you choose the double thorwing, as the two type of check require. But in my specific case $client->post() return a json even if the status code is "500". — EmaOnTheBlock, Jul 30 '14 at 22:12
indeed the json contain the message that explain the error so i need to bypass any guzzle client exception! :( — EmaOnTheBlock, Jul 30 '14 at 22:15

How can I make Symfony2 ignore Guzzle Client bad response exception in my custom controller?

2 Answers2