标签: guzzlehttp

是否有可能让Guzzle池等待请求？

现在我可以动态地向池中添加请求,但是一旦池为空,guzzle就会停止(显然).

当我同时处理10个左右的页面时,这是一个问题,因为我的请求数组将为空,直到处理结果HTML页面并添加新链接.

这是我的发电机:

$generator = function () {
  while ($request = array_shift($this->requests)) {
    if (isset($request['page'])) {
      $key = 'page_' . $request['page'];
    } else {
      $key = 'listing_' . $request['listing'];
    }

    yield $key => new Request('GET', $request['url']);                                          
  }
  echo "Exiting...\n";
  flush();
};

我的游泳池:

$pool = new Pool($this->client, $generator(), [
  'concurrency' => function() {
    return max(1, min(count($this->requests), 2));
  },
  'fulfilled' => function ($response, $index) {
      // new requests may be added to the $this->requests array here
  }
  //...
]);

$promise = …

当我使用Postman进行API调用时,我会收到一个JSON对象.这就是我所期望的.

但是当我像Guzzle一样打电话时:

$client = new \GuzzleHttp\Client(['base_uri' => 'https://api.dev/']);

$response = $client->request('GET', 'search', [
    'verify' => false,
]);

var_dump($response->getBody()); //null

var_dump($response); //returns below

我从Guzzle下面转储

Response {#304 ?
  -reasonPhrase: "OK"
  -statusCode: 200
  -headers: array:8 [?
    "Server" => array:1 [?]
    "Content-Type" => array:1 [?
      0 => "application/json"
    ]
    "Transfer-Encoding" => array:1 [?]
    "Connection" => array:1 [?]
    "Cache-Control" => array:1 [?]
    "Date" => array:1 [?]
    "X-RateLimit-Limit" => array:1 [?]
    "X-RateLimit-Remaining" => array:1 [?]
  ]
  -headerNames: array:8 [?
    "server" => "Server"
    "content-type" => "Content-Type"
    "transfer-encoding" …

我正在尝试为名为creditguard的本地网关创建一个自定义的omnipay驱动程序.对于此网关,您需要将数据发布到端点并返回付款表单的重定向URL.

我的问题是你如何在购买前发布并获得回复？

编辑:

Gateway.php

    class Gateway extends AbstractGateway
{
    public function getName()
    {
        return 'Creditguard';
    }

    public function getDefaultParameters()
    {
        return array();

    }

    public function getEndpoint()
    {
        return 'https://verifonetest.creditguard.co.il/xpo/Relay';
    }



    public function purchase(array $parameters = array())
    {
       return $this->createRequest('\Nirz\Creditguard\Message\PurchaseRequest', $parameters);

    }

    public function completePurchase(array $parameters = array())
    {
        return $this->createRequest('\Nirz\Creditguard\Message\CompletePurchaseRequest', $parameters);
    }

}

PurchaseRequest.php

  class PurchaseRequest extends AbstractRequest
{
    protected $liveEndpoint = 'https://verifonetest.creditguard.co.il/xpo/Relay';
    protected $testEndpoint = 'https://verifonetest.creditguard.co.il/xpo/Relay';


    public function getData()
    {
        $this->validate('amount');

        // Either the nodifyUrl or the returnUrl can …

有没有办法在Guzzle中模拟响应和请求？

我有一个发送一些请求的类,我想测试.

在Guzzle 文档中,我找到了一种方法,如何模拟响应和单独请求.但我怎样才能将它们结合起来？

因为,如果使用历史堆栈,guzzle试图发送一个真实的请求.和签证一样,当我模拟响应处理程序无法测试请求时.

class MyClass {

     public function __construct($guzzleClient) {

        $this->client = $guzzleClient;

    }

    public function registerUser($name, $lang)
    {

           $body = ['name' => $name, 'lang' = $lang, 'state' => 'online'];

           $response = $this->sendRequest('PUT', '/users', ['body' => $body];

           return $response->getStatusCode() == 201;        
    }

   protected function sendRequest($method, $resource, array $options = [])
   {

       try {
           $response = $this->client->request($method, $resource, $options);
       } catch (BadResponseException $e) {
           $response = $e->getResponse();
       }

       $this->response = $response;

      return $response;
  }

}

测试:

class MyClassTest …

我需要集成一个API，所以我要编写函数：

public function test() {

    $client = new GuzzleHttp\Client();

try {
    $res = $client->post('http://example.co.uk/auth/token', [

    'headers' => [
        'Content-Type' => 'application/x-www-form-urlencoded',
            ],

    'json' => [
        'cliend_id' => 'SOMEID',
        'client_secret' => '9999jjjj67Y0LBLq8CbftgfdreehYEI=',
        'grant_type' => 'client_credentials'
]
            ]);

$res = json_decode($res->getBody()->getContents(), true);
dd($res);

}
catch (GuzzleHttp\Exception\ClientException $e) {
        $response = $e->getResponse();
        $result =  json_decode($response->getBody()->getContents());

    return response()->json(['data' => $result]);

    }

}

作为回应，我收到消息：

{"data":{"error":"invalid_clientId","error_description":"ClientId should be sent."}}

现在，当我尝试在POSTMAN应用程序中使用相同的数据运行相同的URL时，我得到了正确的结果：

我的代码有什么不好？我form_params也发送正确的尝试将form_params更改为json，但再次出现相同的错误...

如何解决我的问题？

我试图卷曲到我的本地主机 laravel 站点。我正在使用 XAMPP 7.1 当我尝试重新连接此代码时，总是出现错误（而不是200 OK响应）

致命错误：未捕获的 GuzzleHttp\Exception\ConnectException：cURL 错误 6：无法解析主机：test.example.localhost（参见http://curl.haxx.se/libcurl/c/libcurl-errors.html) 在 D:\xamp7.1\htdocs\wittymanager\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php:185 堆栈跟踪：#0 D:\xamp7.1\htdocs\wittymanager\vendor\guzzlehttp\guzzle\ src\Handler\CurlFactory.php(149): GuzzleHttp\Handler\CurlFactory::createRejection(Object(GuzzleHttp\Handler\EasyHandle), Array) #1 D:\xamp7.1\htdocs\wittymanager\vendor\guzzlehttp\guzzle\ src\Handler\CurlFactory.php(102): GuzzleHttp\Handler\CurlFactory::finishError(Object(GuzzleHttp\Handler\CurlHandler), Object(GuzzleHttp\Handler\EasyHandle), Object(GuzzleHttp\Handler\CurlFactory)) #2D :\xamp7.1\htdocs\wittymanager\vendor\guzzlehttp\guzzle\src\Handler\CurlHandler.php(43): GuzzleHttp\Handler\CurlFactory::finish(Object(GuzzleHttp\Handler\CurlHandler), Object(GuzzleHttp\Handler) \EasyHandle), 对象(GuzzleHttp\Handler\CurlFactory)) #3 D:\xamp7。1\htdocs\wittymanager\vendor\guzzlehttp\guzzle in D:\xamp7.1\htdocs\wittymanager\vendor\guzzlehttp\guzzle\src\Handler\CurlFactory.php on line 185

 // URL = http://test.example.localhost/api/lang-sync
 public static function test($url, $data) {   
    $client = new \GuzzleHttp\Client();

    $res = $client->request('POST', $url, $data);

    $response = $res->getBody();

    if($res->getStatusCode() == 200) 
    {
        print_r($res);

    } else {

        print_r($response);

    }
 }

在登台服务器上没问题。重新启动 xampp 没有帮助。

编辑：

test.example.localhost 已添加到主机文件中。 …

我已经安装了客户端,并使用composer dump autoload进行了更新,但我仍然遇到了同样的错误.在通过composer安装后,需要在项目目录中使用guzzlehttp/guzzle:~6.0.

 $client = new GuzzleHttp\Client(); 

为什么它没有工作,为什么它甚至引用了错误的目录？

我正在构建一个小应用程序Laravel 5.5,我正在使用它Guzzle Http来调用api url并获得响应.很少有api调用具有某些条件,有标题可以作为生成请求的授权.我正在尝试将标题放置如下:

public function post(Request $request)
{
    try {
        if ($request->url_method == 'get') {
            $request = $this->client->get($request->url, [ 'headers' => [ 'access_token' => $request->headers->access_token]]);
        }
        else if ($request->url_method == 'post')
        {  $request = $this->client->post($request->url, [$request->request_data]);  }
        else { return response()->json(['data' => 'Invalid Method'],500);   }

        $response = \GuzzleHttp\json_decode($request->getBody());
        return response()->json(['data' => json_decode($response->d)], $request->getStatusCode());
    }
    catch (ClientException $e) {
        return response()->json(['data' => 'Invalid Request.'], $request->getStatusCode());
    }
}

但这给了我错误:

未定义的属性:Symfony\Component\HttpFoundation\HeaderBag :: $ access_token

请查看截图:

此外,当在控制台中通过浏览器进行调用时,它给出了同样的错误:

请帮帮我,谢谢.

Guzzle http 正在截断超过 120 个字符的异常，但我需要记录完整的异常消息。我怎样才能做到这一点？

我正在使用 Laravel 4.2.22。

我尝试GuzzleHttp在本地使用并向本地 URL 发出请求，但收到了Operation timed out. 使用的代码片段：

    $url = 'https://boot-vue.test'; //local domain, tried with other urls as well
    $client = new \GuzzleHttp\Client([
        'verify' => false,
        'timeout' => 5, // Response timeout
        'connect_timeout' => 5, // Connection timeout
        'peer' => false
    ]);
    $response = $client->request($method, $url, [
        'json' => $data,
        'headers' => $headers,
    ]);

上面的代码段适用于任何公共网站。

任何帮助都是可观的！