Play Framework 2.0 如何在 Java 中使用 OpenID？

Question

如何在 Java 中使用 Play!2.0 OpenID？

我找到了一个example，但我收到了这个错误：

! @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]

documentation about OpenID in Java 不够用。

Answer 1

试试我的示例https://gist.github.com/2889927