file_get_contents('php:// input')保持返回空

Question

我正在尝试在我的网站上设置gitkit,但无法通过这一行代码.无论我做什么,都会file_get_contents一直空着.

我已经设置了我的php.ini: always_populate_raw_post_data = On

我的环境是PHP 5.3.3,Apache 2.2.6为localhost.

这是一些代码.

在我的index.php中,我调用google API并尝试使用gmail帐户登录,换句话说,联合登录.

(这是来自Google API控制台)

<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.min.js"></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/jqueryui/1.8.2/jquery-ui.min.js"></script>
<script type="text/javascript" src="//ajax.googleapis.com/ajax/libs/googleapis/0.0.4/googleapis.min.js"></script>
<script type="text/javascript" src="//ajax.googleapis.com/jsapi"></script>
<script type="text/javascript">
  google.load("identitytoolkit", "1", {packages: ["ac"], language:"en"});
</script>
<script type="text/javascript">
  $(function() {
    window.google.identitytoolkit.setConfig({
        developerKey: "HERE_GOES_MY_DEVELOPER_KEY",
        companyName: "Valentinos Pizzaria",
        callbackUrl: "http://localhost/valentinos/callback.php",
        realm: "",
        userStatusUrl: "http://localhost/valentinos/userstatus.php",
        loginUrl: "http://localhost/valentinos/login.php",
        signupUrl: "http://localhost/valentinos/register.php",
        homeUrl: "http://localhost/valentinos/index.php",
        logoutUrl: "http://localhost/valentinos/logout.php",
        idps: ["Gmail", "Yahoo"],
        tryFederatedFirst: true,
        useCachedUserStatus: false,
        useContextParam: true
    });
    $("#navbar").accountChooser();
  });
</script>

我收到IDP响应,登录,并被要求获得权限.返回我的回调页面后,我使用了Google提供的代码示例(如下所示),这一行代码似乎没有正确返回.

我做了什么愚蠢的事吗？

任何帮助将不胜感激.

到目前为止,这里是整个callback.php(现在没有任何HTML):

  session_start();

  $url = EasyRpService::getCurrentUrl();
  #$postData = @file_get_contents('php://input');
  $postData = file_get_contents('php://input');
  $result = EasyRpService::verify($url, $postData);
  // Turn on for debugging.
  // var_dump($result);

class EasyRpService {
  // Replace $YOUR_DEVELOPER_KEY
  private static $SERVER_URL = 'https://www.googleapis.com/rpc?key=HERE_GOES_MY_DEVELOPER_KEY';

  public static function getCurrentUrl() {
    $url = (isset($_SERVER['HTTPS']) && $_SERVER['HTTPS'] == 'on') ? 'https://' : 'http://';
    $url .= $_SERVER['SERVER_NAME'];
    if ($_SERVER['SERVER_PORT'] != '80') {
      $url .= ':'. $_SERVER['SERVER_PORT'];
    }
    $url .= $_SERVER['REQUEST_URI'];
    return $url;
  }

  private static function post($postData) {
    $ch = curl_init();
    curl_setopt_array($ch, array(
        CURLOPT_URL => EasyRpService::$SERVER_URL,
        CURLOPT_RETURNTRANSFER => 1,
        CURLOPT_HTTPHEADER => array('Content-Type: application/json'),
        CURLOPT_POSTFIELDS => json_encode($postData)));
    $response = curl_exec($ch);
    $http_code = curl_getinfo($ch, CURLINFO_HTTP_CODE);
    curl_close($ch);
    if ($http_code == '200' && !empty($response)) {
      return json_decode($response, true);
    }
    return NULL;
  }

  public static function verify($continueUri, $response) {
    $request = array();
    $request['method'] = 'identitytoolkit.relyingparty.verifyAssertion';
    $request['apiVersion'] = 'v1';
    $request['params'] = array();
    $request['params']['requestUri'] = $continueUri;
    $request['params']['postBody'] = $response;

    $result = EasyRpService::post($request);
    if (!empty($result['result'])) {
      return $result['result'];
    }
    return NULL;
  }

} # End Class EasyRpService

在有人要求之前,我会HERE_GOES_MY_DEVELOPER_KEY用我的开发人员密钥替换...

再一次,任何帮助将不胜感激.C ya.

Answer 1

你尝试过使用$ _POST吗？php://input 不适用于enctype ="multipart/form-data".在这种情况下,$ _POST [0]可能会起作用,因为你可以获得多部分/表格数据的响应.