Play Framework 2.0如何在Java中使用OpenID？

Question

如何在Java中使用Play！2.0 OpenID？

我找到了一個例子 ，但是我收到了這個錯誤：

! @6af3een21 - Internal server error, for request [GET /login/verify] ->

play.core.ActionInvoker$$anonfun$receive$1$$anon$1: Execution exception [[Errors$BAD_RESPONSE$: null]]
at play.core.ActionInvoker$$anonfun$receive$1.apply(Invoker.scala:82) [play_2.9.1.jar:2.0]
at play.core.ActionInvoker$$anonfun$receive$1.apply(Invoker.scala:63) [play_2.9.1.jar:2.0]
at akka.actor.Actor$class.apply(Actor.scala:290) [akka-actor.jar:2.0]
at play.core.ActionInvoker.apply(Invoker.scala:61) [play_2.9.1.jar:2.0]
at akka.actor.ActorCell.invoke(ActorCell.scala:617) [akka-actor.jar:2.0]
at akka.dispatch.Mailbox.processMailbox(Mailbox.scala:179) [akka-actor.jar:2.0]
Caused by: play.api.libs.openid.Errors$BAD_RESPONSE$: null
at play.api.libs.openid.Errors$BAD_RESPONSE$.<clinit>(OpenIDError.scala) ~[play_2.9.1.jar:2.0]
at play.api.libs.openid.OpenID$$anonfun$verifiedId$7.apply(OpenID.scala:88) ~[play_2.9.1.jar:2.0]
at play.api.libs.openid.OpenID$$anonfun$verifiedId$7.apply(OpenID.scala:88) ~[play_2.9.1.jar:2.0]
at scala.util.control.Exception$Catch$$anonfun$either$1.apply(Exception.scala:110) ~[scala-library.jar:na]
at scala.util.control.Exception$Catch$$anonfun$either$1.apply(Exception.scala:110) ~[scala-library.jar:na]
at scala.util.control.Exception$Catch.apply(Exception.scala:88) ~[scala-library.jar:na]

有關Java中OpenID的文檔是不夠的。

Answer 1

試試我的樣本https://gist.github.com/2889927