如何返回Guzzle JSON响应 [英] How to return Guzzle JSON response

问题描述

我正在使用Guzzle发出一个返回JSON的aSync请求.通话正常，响应正常，但是:

I am using Guzzle to make a aSync request that returns JSON. The call is working fine and the response is ok, however:

$client = new Client();
    $promise = $client->requestAsync($requestType ,$this->url.$resource, // endpoint
        [
            'auth' => [ // credentials
                $this->username, 
                $this->password
            ],
            'json' => $payload, // the package
            'curl' => [ // some curl options
                CURLOPT_HTTPAUTH => CURLAUTH_BASIC,
                CURLOPT_RETURNTRANSFER => true,
            ],
            'headers' => [ // custom headers
                'Accept' =>  'application/json',
                'Content-Type' => 'application/json'
            ]
        ]
    );

    $response = $promise->wait();
    echo $response->getStatusCode().'<br /><br />';
    // Error handling
    if($response->getStatusCode() != 200){
        // Error Handling
    }else{
        echo $response->getBody(true);
    }

如果我回显response-> getBody()，我会看到JSON字符串，但是如果将其分配给属性print_r或返回它，则会得到:

if I echo response->getBody() I see the JSON string, but if I assign it to a property, print_r, or return it I get:

GuzzleHttp\Psr7\Stream Object ( [stream:GuzzleHttp\Psr7\Stream:private] => Resource id #245 [size:GuzzleHttp\Psr7\Stream:private] => [seekable:GuzzleHttp\Psr7\Stream:private] => 1 [readable:GuzzleHttp\Psr7\Stream:private] => 1 [writable:GuzzleHttp\Psr7\Stream:private] => 1 [uri:GuzzleHttp\Psr7\Stream:private] => php://temp [customMetadata:GuzzleHttp\Psr7\Stream:private] => Array ( ) )

我需要使用JSON来验证我对服务的响应.我怎样才能做到这一点?我已经阅读了文档，但是显然我遗漏了一些东西.

I need to use the JSON to validate my response from the service. How can I do this? I have gone through the docs, but am I obviously missing something.

基本上是在将json getBody输出分配为$ json的过程中:

Essentially along the lines of assigning the json getBody output to say $json:

if($json->first_field > 0)

任何帮助表示赞赏. 问候

Any Help appreciated. Regards

推荐答案

在对SO进行了更多研究之后，我首先跌入了这篇文章

After some more research on SO I tumbled head first into this post

枪口6:不再有json()方法寻求回应

基本上，执行以下操作将返回原始输出.

Essentially doing the following will return the raw output.

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

巨大的头痛消失了.希望这对某人有帮助

Huge headache gone. Hope this helps someone

这篇关于如何返回Guzzle JSON响应的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！