Play Framework 2.0如何在Java中使用OpenID？

Question

How can I use Play!2.0 OpenID in Java?

I found an example , but I get this error:

! @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]

The documentation about OpenID in Java is not sufficient.

Answer 1

试试我的样本https://gist.github.com/2889927