Question

使用LightOpenID课程，我正在尝试将Google的登录信息与我的网站集成。我正在使用默认代码：

<?php
require 'includes/openid.php';

try {
    if(!isset($_GET['openid_mode'])) {
        if(isset($_GET['login'])) {
            $openid = new LightOpenID;
            $openid->identity = 'https://www.google.com/accounts/o8/id';
            header('Location: ' . $openid->authUrl());
        }
?>
<form action="?login" method="post">
    <button>Login with Google</button>
</form>
<?php
    } elseif($_GET['openid_mode'] == 'cancel') {
        echo 'User has canceled authentication!';
    } else {
        $openid = new LightOpenID;
        echo 'User ' . ($openid->validate() ? $openid->identity . ' has ' : 'has not ') . 'logged in.';
    }
} catch(ErrorException $e) {
    echo $e->getMessage();
}
echo '<pre>'.print_r($openid,true).'</pre>';
?>

最初，这告诉我我缺少构造方法的参数1，迫使我将new LightOpenID方法称为new LightOpenID($host)。但是因为我已经改变它并尝试登录，所以当我恢复代码时它没有给出错误只是去抓错误...

无论如何，除此之外，在我添加主机后，我被带到谷歌页面，询问我是否想用我的谷歌帐户登录该网站。我按下了登录按钮并被收回，但被告知我没有登录。

User has not logged in.

LightOpenID Object
(
    [returnUrl] => myReturnURL
    [required] => Array
        (
        )

    [optional] => Array
        (
        )

    [verify_peer] => 
    [capath] => 
    [cainfo] => 
    [data] => Array
        (
            [login] => 
            [openid_ns] => http://specs.openid.net/auth/2.0
            [openid_mode] => id_res
            [openid_op_endpoint] => https://www.google.com/accounts/o8/ud
            [openid_response_nonce] => 2012-07-23T14:06:24ZQwlS1zWFRdvg6A
            [openid_return_to] => myReturnURL
            [openid_assoc_handle] => AMlYA9WMwp_rTkH3swNLEF5UASFvhyOR-kqVSXGe7cvXe7ws5z8HHWUu
            [openid_signed] => op_endpoint,claimed_id,identity,return_to,response_nonce,assoc_handle
            [openid_sig] => NtOEY8dUM1Hpt2tTR9x0RdN5o1c=
            [openid_identity] => https://www.google.com/accounts/o8/id?id=AItOawkXNCi5qq-3Vj_esgSQAwmEN_nVIC8BHAQ
            [openid_claimed_id] => https://www.google.com/accounts/o8/id?id=AItOawkXNCi5qq-3Vj_esgSQAwmEN_nVIC8BHAQ
        )

    [identity:LightOpenID:private] => 
    [claimed_id:LightOpenID:private] => https://www.google.com/accounts/o8/id?id=AItOawkXNCi5qq-3Vj_esgSQAwmEN_nVIC8BHAQ
    [server:protected] => https://www.google.com/accounts/o8/ud
    [version:protected] => 2
    [trustRoot:protected] => mySite
    [aliases:protected] => 
    [identifier_select:protected] => 
    [ax:protected] => 1
    [sreg:protected] => 
    [setup_url:protected] => 
    [headers:protected] => Array
        (
            [cache-control] => no-cache, no-store, max-age=0, must-revalidate
            [pragma] => no-cache
            [expires] => Fri, 01 Jan 1990 00:00:00 GMT
            [date] => Mon, 23 Jul 2012 14:06:26 GMT
            [content-type] => text/plain; charset=UTF-8
            [x-content-type-options] => nosniff
            [x-frame-options] => SAMEORIGIN
            [x-xss-protection] => 1; mode=block
            [server] => GSE
        )

)

我已经登录Google，因此Google页面没有要求我登录，只是为了确认我想使用我的Google帐户登录我自己的网站。有没有人看到为什么不说我已登录？

提前致谢。

Answer 1

使用newest version，它应该有效。

请记住，较新版本的库需要构造函数中的参数：

$openid = new LightOpenId('www.domain.com');

LightOpenID使用默认设置提供错误，并使用有效登录返回false

1 个答案: