curl - 将 paypal curl get access token 请求转换为 guzzle

Question

我需要一些关于 paypal rest api 的帮助。

我使用 guzzle 作为 http 客户端来使用 paypal api

当我在命令行中使用 curl 尝试 paypal 示例时，它确实有效

但是当我想用 guzzle 重现它时，我总是从 paypal 得到 Internal 500 ERROR..

这是官方文档中的 paypal curl 示例(请在此处查看 https://developer.paypal.com/webapps/developer/docs/integration/direct/make-your-first-call/):

curl -v https://api.sandbox.paypal.com/v1/oauth2/token \
  -H "Accept: application/json" \
  -H "Accept-Language: en_US" \
  -u "clientId:clientSecret" \
  -d "grant_type=client_credentials"

这是我的 Guzzle 代码:

/**
 * Etape 1 récuperer un access token
 */
$authResponse = $client->get("https://api.sandbox.paypal.com/v1/oauth2/token", [
    'auth' =>  [$apiClientId, $apiClientSecret, 'basic'],
    'body' => ['grant_type' => 'client_credentials'],
    'headers' => [
    'Accept-Language' => 'en_US',
    'Accept'     => 'application/json' 
    ]
]);

echo $authResponse->getBody();

我尝试过 auth basic、digest，但到目前为止都没有用。

感谢您对此的任何帮助!

Answer 1

这是使用 guzzlehttp

$uri = 'https://api.sandbox.paypal.com/v1/oauth2/token';
$clientId = \\Your client_id which you got on sign up;
$secret = \\Your secret which you got on sign up;

$client = new \GuzzleHttp\Client();
$response = $client->request('POST', $uri, [
        'headers' =>
            [
                'Accept' => 'application/json',
                'Accept-Language' => 'en_US',
               'Content-Type' => 'application/x-www-form-urlencoded',
            ],
        'body' => 'grant_type=client_credentials',

        'auth' => [$clientId, $secret, 'basic']
    ]
);

$data = json_decode($response->getBody(), true);

$access_token = $data['access_token'];