Guzzle 6: no more json() method for responses

call to undefined method guzzlehttp\psr7\response::json()
guzzle 6 json response
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Previously in Guzzle 5.3:

$response = $client->get('http://httpbin.org/get');
$array = $response->json(); // Yoohoo
var_dump($array[0]['origin']);

I could easily get a PHP array from a JSON response. Now In Guzzle 6, I don't know how to do. There seems to be no json() method anymore. I (quickly) read the doc from the latest version and don't found anything about JSON responses. I think I missed something, maybe there is a new concept that I don't understand (or maybe I did not read correctly).

Is this (below) new way the only way?

$response = $client->get('http://httpbin.org/get');
$array = json_decode($response->getBody()->getContents(), true); // :'(
var_dump($array[0]['origin']);

Or is there an helper or something like that?


I use json_decode($response->getBody()) now instead of $response->json().

I suspect this might be a casualty of PSR-7 compliance.

no more json() method for responses? · Issue #1106 · guzzle/guzzle , Call to undefined method GuzzleHttp\Psr7\Response::json() @duncan-brown · json() is deprecated in guzzle 6 … c861f8c · guzzle/guzzle#  This would then lead to needing to incessantly wrap normal PSR-7 responses with Guzzle specific responses or you would need to check the class of each response object you receive from Guzzle to ensure that it has the json method. Because Guzzle has a middleware system that is meant to essentially decorate clients, it makes much more sense to


You switch to:

json_decode($response->getBody(), true)

Instead of the other comment if you want it to work exactly as before in order to get arrays instead of objects.

Using Response objects, Sending a request will return a Guzzle\Http\Message\Response object. You can See HTTP Message Headers for more information. You can easily parse and use a JSON response as an array using the json() method of a response. You will get an exception if you call this method and the response is not valid JSON. The Guzzle response indicates a 200 status code on the request, so I'm not sure exactly what I need to do to retrieve the returned data. Guzzle implements PSR-7 . That means that it will by default store the body of a message in a Stream that uses PHP temp streams.


I use $response->getBody()->getContents() to get JSON from response. Guzzle version 6.3.0.

Request and Response Messages, Both request and response messages contain HTTP headers. Header values that have no key (i.e., the link) are indexed numerically while More precisely, clients use a GuzzleHttp\Message\MessageFactoryInterface to You create requests with a client using the createRequest() method. $json = $response->​json();. To resolve the “the response is not a valid JSON response” or mixed content error, navigate to Settings > SSL. This will open the plugin’s settings. This will open the plugin’s settings. Now click on the Settings tab.


If you guys still interested, here is my workaround based on Guzzle middleware feature:

  1. Create JsonAwaraResponse that will decode JSON response by Content-Type HTTP header, if not - it will act as standard Guzzle Response:

    <?php
    
    namespace GuzzleHttp\Psr7;
    
    
    class JsonAwareResponse extends Response
    {
        /**
         * Cache for performance
         * @var array
         */
        private $json;
    
        public function getBody()
        {
            if ($this->json) {
                return $this->json;
            }
            // get parent Body stream
            $body = parent::getBody();
    
            // if JSON HTTP header detected - then decode
            if (false !== strpos($this->getHeaderLine('Content-Type'), 'application/json')) {
                return $this->json = \json_decode($body, true);
            }
            return $body;
        }
    }
    
  2. Create Middleware which going to replace Guzzle PSR-7 responses with above Response implementation:

    <?php
    
    $client = new \GuzzleHttp\Client();
    
    /** @var HandlerStack $handler */
    $handler = $client->getConfig('handler');
    $handler->push(\GuzzleHttp\Middleware::mapResponse(function (\Psr\Http\Message\ResponseInterface $response) {
        return new \GuzzleHttp\Psr7\JsonAwareResponse(
            $response->getStatusCode(),
            $response->getHeaders(),
            $response->getBody(),
            $response->getProtocolVersion(),
            $response->getReasonPhrase()
        );
    }), 'json_decode_middleware');
    

After this to retrieve JSON as PHP native array use Guzzle as always:

$jsonArray = $client->get('http://httpbin.org/headers')->getBody();

Tested with guzzlehttp/guzzle 6.3.3

Quickstart, Clients are immutable in Guzzle 6, which means that you cannot change the defaults used You can find out more about client middleware in the Handlers and You can also use the sendAsync() and requestAsync() methods of a client: if any of the requests fail $responses = Promise\unwrap($promises); // Wait for the  45. For Guzzle <= 4: It's a raw post request so putting the JSON in the body solved the problem. $request = $this->client->post($url,array( 'content-type' => 'application/json' ),array()); $request->setBody($data); #set body! $response = $request->send(); return $response; share.


Adding ->getContents() doesn't return jSON response, instead it returns as text.

You can simply use json_decode

JSON Responses in Guzzle 6, Guzzle is a wildly popular PHP client that makes it simple to perform One small side effect of this was the removal of the "::json()" method from the response, as the PSR-7 standard does not define this So much more than just the actual body. I could add another method, say, "::jsonSerialize()" from the  When no more responses are in the queue and a request is sent, an OutOfBoundsException is thrown. History Middleware ¶ When using things like the Mock handler, you often need to know if the requests you expected to send were sent exactly as you intended.


Guzzle 6: no more json() method for responses, Guzzle 6: no more json() method for responses. likeanswer Previously in Guzzle 5.3: $response = $client->get('http://httpbin.org/get'); $array  Laravel API Errors and Exceptions: How to Return Responses February 13, 2019 API-based projects are more and more popular, and they are pretty easy to create in Laravel.


[PDF] Guzzle, You can more easily work with JSON responses using the json() method of a response. $response = $client->get('http://httpbin.org/get');. Guzzle¶. Guzzle is a PHP HTTP client that makes it easy to send HTTP requests and trivial to integrate with web services. Manages things like persistent connections, represents query strings as collections, simplifies sending streaming POST requests with fields and files, and abstracts away the underlying HTTP transport layer.


Handling guzzle 6 response body in laravel, I've already read the Guzzle 6 response body documentation, but it's not that about JSON and Guzzle here: Guzzle 6: no more json() method for responses  When no more responses are in the queue and a request is sent, an OutOfBoundsException is thrown. History Middleware ¶ When using things like the Mock handler, you often need to know if the requests you expected to send were sent exactly as you intended.