Question

尝试使用此Google客户端API示例

https://developers.google.com/identity/sign-in/web/backend-auth

我通过javascript获取访问令牌并通过ajax javascript对象将其发布到我的PHP。

expires_at
:
1490552860336
expires_in
:
2223
first_issued_at
:
1490549260336
id_token
:
"eyJhbGciOiJSUzI1NiIsImtpZCI6ImRkMDM4NzVmMzkzMTdiZThlNTc0MDBlNmMzMDYzMmFmNTU4YzJkZTQifQ.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.PBcvfMs9_MO9mcHDtGGyb3VbO-O374tHH7w4gWr8ZMpRDYyynY6ZQd5BxRxlh6b__upzxz6gMYQ29yY__zDnEp4_NrkSp7RWQAcoF-JuY8xM1GtDy5IwEFKPxDh9h-9bGoxCx5jgX7-bu2F5Xk6opwZUzoyyYk5BT1xXpBfAe35CJHK_NAzSfrwEX8vU-hIP9C_BjipXptg5Eo3fi3DxPWRVbrzsscV0DDASm9xEi5ilLkUxBwRZN9bk6hgXXgn2Ts7Mo4B7Rue1oyB_rkNJ3RSgH7ylnHrVePiVwJ_15aKwiD64FgcmdLOruLEm9pADoiAodMVNZ6HAElr3MuWG1w"
idpId
:
"google"
login_hint
:
"AJDLj6JimYVJadfBfXai38izEPbdRaSSCtY5FyzYIVQ1IPE2kad13fq5-9Qwkik2AlWx3_Qlco531-R7LVEwcxvs8Xhw0TvrSg"
session_state
:
Object
token_type
:
"Bearer"

我的PHP如下

  require_once 'google-api-php-client-2.1.1/vendor/autoload.php';
            // Get $id_token via HTTPS POST.

            $client = new Google_Client(['client_id' =>'my google client id' ]);
            $payload = $client->verifyIdToken($this->input->post('id_token'));
            if ($payload) {
              $userid = $payload['sub'];
              // If request specified a G Suite domain:
              //$domain = $payload['hd'];
            } else {
              // Invalid ID token
            }

令人沮丧的是，我从来没有幸运过谷歌API示例。我似乎无法在网上找到任何可以让它在这些日子里工作的东西。请帮我。我基本上都在寻找可以验证并保存在我的数据库中的ID或令牌

我被困在这个功能上

$payload = $client->verifyIdToken($this->input->post('id_token'));

屏幕截图显示了在抛出错误之前我得到了多远。

Screen shot of error

我最终走了这条路线

  $q = 'https://www.googleapis.com/oauth2/v3/tokeninfo?id_token=' . $this->input->post('id_token');
            $json = file_get_contents($q);
            $userInfoArray = json_decode($json,true);
            $googleEmail = $userInfoArray['email'];
            $google_id= $userInfoArray['kid'];

我不确定该孩子是否是该用户的唯一ID，但它不断返回相同的值。

Answer 1

感谢，我的

<?php
require_once 'google-api-php-client-2.2.0_PHP54/vendor/autoload.php';

// Get $id_token via HTTPS POST.
$id_token= $_POST['id_token'];
$CLIENT_ID = 'yourclientid......apps.googleusercontent.com';
$client = new Google_Client(['client_id' => $CLIENT_ID]);
$payload = 'https://www.googleapis.com/oauth2/v3/tokeninfo?id_token=' . $id_token;
        $json = file_get_contents($payload);
        $userInfoArray = json_decode($json,true);
        $googleEmail = $userInfoArray['email'];
        $google_id= $userInfoArray['sub'];

if ($googleEmail)
 {
  echo "validated".$googleEmail;
} else {
   echo "inValidate" ;
  }
?>

google client api使用无效的后端服务器进行身份验证

1 个答案: