如何从带有'；的url中检索参数'；

在用户通过openID登录谷歌后，我正在尝试检索用户的电子邮件地址。

网址是：

00.000000.000/使用谷歌.php登录？openid.ns=http%3A%2F%2Specifics.openid.net%2Fauth%2F2.0&openid.mode=id_res&openid.op_endpoint=https%3A%2F%2Fwww.google.com%2Faccounts%2Fo8%2Fud&openid.response_nonce=2013-08-12T20%3A52%3A27ZoZizKCa486SFiQ&openid=return_to=http%3A%2F%2F24.255.213.250%3A50005%2Floginwithgoogle.php&openid=invalidate_handle=ABSmpf6DNMw&openid.assoc_handle=1.AMlYA9XT63izbhULzG8CiL5xkIE9bfgIV6dQ_5xBjHZQjVnh4H5YRm4L2HSTRxYj&openid.signed=op_endpoint%2claimed_id%2identity%2Return_to%2response_none%2assoc_handle%2Cns.ext1%2Cext1.mode%2ext1.type.email%2ext1.value.email&openid=xK06wAKPUPdu4JVgLz0v%2F1ZTMZA%3D&openid.identity=https%3A%2F%2Fwww.google.com%2Faccounts%2Fo8%2Fid%3Id%3DAITOWm70UATPuqUjklL10scHQJGXvepPFSMI48&openid=claimed_id=https%3A%2F%2Fwww.google.com%2Faccounts%2Fo8%2Fid%3Id%3DAITOWm70UATPuqUjklL10scHQJGXvepPFSMI48&openid.ns.ext1=http%3A%2F%2Openid.net%2Fsrv%2Fax%2F1.0&openid.ext1.mode=获取响应&openid.ext1.type.email=http%3A%2F%2Faxschema.org/2Fcontact%2Femail&openid=ext1.value.email=myemail%40gmail.com

php代码是：

<?php
    echo $_POST["openid.ext1.value.email"];
    echo $_GET["openid.ext1.value.email"];
?>

没有显示任何内容。PHP和get函数都在工作。我认为问题是试图用"获取参数（期间）。我想要的只是电子邮件，所以如果我用一种过于复杂的方式来做这件事，我会很高兴听到其他选择。提前谢谢。