Question

假设我想在获取授权代码后编写一个php命令来兑换授权代码。

假设我使用guzzle来发出请求，我正在使用以下代码，其中参数应该是正确的。

<?php

namespace Pherserk\PHPGMailCommander\Command;
use Guzzle\Http\Client;

/**
 * Class ExchangeAuthorizationToken
 * @package Pherserk\PHPGmailCommander\Command
 */
class ExchangeAuthorizationToken
{
    private $url;
    private $path;
    private $grantType;
    private $clientId;
    private $clientSecret;
    private $redirectUri;

    public function __construct(array $config)
    {
        $this->url = $config['authorization_token']['code_exchange_uri'];
        $this->path = $config['authorization_token']['code_exchange_path'];
        $this->grantType = $config['authorization_token']['grant_type'];
        $this->clientId = $config['client_id'];
        $this->clientSecret = $config['client_secret'];
        $this->redirectUri = $config['redirect_uri'];
    }

    /**
     * @param string $accessToken
     */
    public function run($accessToken = '4/the-authorization-token-here_hasbEen.obscured')
    {
        $client = new Client();
        $client->setBaseUrl($this->url);
        $client->setSslVerification(true);

        $request = $client->post($this->path, null, null, ['exceptions' => false]);
        $request->addHeader('Content-Type', 'application/x-www-form-urlencoded');

        $query = $request->getQuery();
        $query['code'] = $accessToken;
        $query['client_id'] = $this->clientId;
        $query['client_secret'] = $this->clientSecret;
        $query['redirect_uri'] = $this->redirectUri;
        $query['grant_type'] = $this->grantType;

        $response = $request->send();
        var_dump($response->getBody(true));
    }
}

我收到400状态代码，动机为invalid_grant。我收到了用户授权提示和重定向后我正在使用的授权码。我错过了什么？

Answer 1

经过一段时间的辩论，我明白了原因。代码已被重新兑换。

以这种方式调整代码

$response = $request->send();
$decodedResponse = $response->json();

if (isset($decodedResponse['error'])) {
    echo $decodedResponse['error_description'];

    return false;
}

return true;

我注意到了错误。

使用带有guzzle的Google API兑换OAuth api access的授权代码

1 个答案: