我正在尝试在我的网站中设置gitkit,但无法通过这一行代码。无论我做什么,file_get_contents
都会一直空着。
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
,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回复,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。
我做了什么蠢事吗?
任何帮助将不胜感激。
到目前为止,这是整个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 $result['result'];
}
return NULL;
}
} # End Class EasyRpService
在有人要求之前,我会用我的开发人员密钥替换HERE_GOES_MY_DEVELOPER_KEY
...
再一次,任何帮助将不胜感激。 C ya。
答案 0 :(得分:6)
您尝试使用$ _POST吗?对于enctype =“multipart / form-data”,php://input
don't work。在这种情况下,$ _POST [0]可能有效,你可能会得到多部分/表格数据的响应。
答案 1 :(得分:3)
HTTP POST数据通常填充在$ _POST中,php://输入通常包含PUT数据。