Question

我意识到这些帖子中大约有20个，但我已经查看了所有这些帖子，到目前为止没有任何答案有帮助（虽然帮助我排除了事情）。

我在PHP中编写了一堆与Youtube相关的api函数，一旦我有访问代码（也没有刷新访问代码的问题），我就没有问题，但是当我尝试交换授权代码时对于访问令牌，我收到“invalid_grant”响应。

我还需要离线访问，因为我计划在没有用户交互的情况下进行api调用，并且不能预先授权帐户。

注意：

我已经使用OAuth游乐场取消了每项测试
我在我的unix系统上设置了“* ntpdate * ”来谷歌
令人耳目一新的令牌作品
如果我从代码中删除 urlencode 并请求uri字符串，则会收到“无效请求”错误。有趣的是它不需要刷新令牌功能。

代码

access.php

$ytclient = new ytClient;

if(isset($_GET['code'])){
  $ytclient->auth_code = $_GET['code'];
  $ytclient->exchangeToken();
}else{
  $ytclient->authorize();
}

ytClient

function authorize(){
  include('config.php');
  echo "<script>window.location = 'https://accounts.google.com/o/oauth2/auth?redirect_uri=".urlencode( $config->domain.'/lib/access.php')."&response_type=code&client_id=**************&scope=https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyt-analytics-monetary.readonly+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyt-analytics.readonly+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyoutube+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyoutube.readonly+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyoutube.upload+https%3A%2F%2Fwww.googleapis.com%2Fauth%2Fyoutubepartner&approval_prompt=force&access_type=offline'</script>";
}
function exchangeToken(){
  $postArr["client_id"] = ($config->ytclientid);
  $postArr["client_secret"] = ($config->ytclientsecret);
  $postArr["code"] = urlencode($this->auth_code);
  $postArr["request_uri"] = urlencode($config->domain."/lib/access.php");
  $postArr["grant_type"] = "authorization_code";
  $accessObj = $this->get_yt_json("https://accounts.google.com/o/oauth2/token", $postArr, 1);
}

function get_yt_json($url, $postArr, $mode){
  if($mode == 1){
    $curl = curl_init($url);
    curl_setopt($curl, CURLOPT_POST, true);
    curl_setopt($curl, CURLOPT_POSTFIELDS, $postArr);
  }elseif($mode == 2){
    $url .= "?".http_build_query($postArr);
    $curl = curl_init($url);
  }
  curl_setopt($curl, CURLOPT_HTTPAUTH, CURLAUTH_ANY);
  curl_setopt($curl, CURLOPT_SSL_VERIFYPEER, false);
  curl_setopt($curl, CURLOPT_HEADER, true);
  curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1);

  $rawJson = curl_exec($curl);
  curl_close($curl);
  echo $rawJson;
  /*
  $obj = json_decode($rawJson);

  return ($obj);
  */
}

响应

HTTP/1.1 400 Bad Request
Cache-Control: no-cache, no-store, max-age=0, must-revalidate 
Pragma: no-cache 
Expires: Fri, 01 Jan 1990 00:00:00 GMT 
Date: Sat, 21 Sep 2013 18:58:31 GMT 
Content-Type: application/json 
X-Content-Type-Options: nosniff 
X-Frame-Options: SAMEORIGIN 
X-XSS-Protection: 1; mode=block 
Server: GSE 
Alternate-Protocol: 443:quic 
Transfer-Encoding: chunked 

{ "error" : "invalid_grant" }

Answer 1

从未找到解决方案，但使用高度无证的google php api =＆gt;修复了它。 https://code.google.com/p/google-api-php-client/

快速简便的解决方案 - 据我所知，魔术......以下是使用api对使用终身令牌验证某人的代码，可能有助于某人：

include(dirname(__FILE__).'/google-api-php-client/src/Google_Client.php');

$client = new Google_Client();
$client->setApprovalPrompt('force');
$client->setAccessType('offline');
$client->setScopes('https://www.googleapis.com/auth/yt-analytics-monetary.readonly https://www.googleapis.com/auth/yt-analytics.readonly https://www.googleapis.com/auth/youtube https://www.googleapis.com/auth/youtube.readonly https://www.googleapis.com/auth/youtube.upload https://www.googleapis.com/auth/youtubepartner');

/* Auth the client and get Token Object */
$auth = $client->authenticate();
$token = json_decode($client->getAccessToken());

PHP Google Api代码交换结果“invalid_grant”

注意：

代码

1 个答案: