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我有Play 2應用程序中運行的AKKA actors。有一個POJO對象列表從數據庫中檢索並傳遞給參與者。當一個actor開始處理這些對象時,它會拋出這個異常。我想它試圖從數據庫讀取數據,因爲ebean的延遲加載。在測試用例中運行時發生這種情況。我沒有在正常的應用程序環境中測試。Play 2.0試圖從已經關機的池中獲取連接
試圖獲得一個池已經被關閉
at com.avaje.ebeaninternal.server.transaction.TransactionManager.createQueryTransaction(TransactionManager.java:356)
at com.avaje.ebeaninternal.server.core.DefaultServer.createQueryTransaction(DefaultServer.java:2021)
at com.avaje.ebeaninternal.server.core.OrmQueryRequest.initTransIfRequired(OrmQueryRequest.java:241)
at com.avaje.ebeaninternal.server.core.DefaultServer.findList(DefaultServer.java:1468)
at com.avaje.ebeaninternal.server.core.DefaultBeanLoader.loadBean(DefaultBeanLoader.java:360)
at com.avaje.ebeaninternal.server.core.DefaultServer.loadBean(DefaultServer.java:526)
at com.avaje.ebeaninternal.server.loadcontext.DLoadBeanContext.loadBean(DLoadBeanContext.java:143)
at com.avaje.ebean.bean.EntityBeanIntercept.loadBean(EntityBeanIntercept.java:548)
at com.avaje.ebean.bean.EntityBeanIntercept.preGetter(EntityBeanIntercept.java:638)
at models.MemberInfo._ebean_get_type(MemberInfo.java:4)
at models.MemberInfo.getType(MemberInfo.java:232)
at actors.MessageWorker.doSendToIOS(MessageWorker.java:161)
at actors.MessageWorker.onReceive(MessageWorker.java:97)
at akka.actor.UntypedActor$$anonfun$receive$1.apply(UntypedActor.scala:154)
at akka.actor.UntypedActor$$anonfun$receive$1.apply(UntypedActor.scala:153)
at akka.actor.Actor$class.apply(Actor.scala:311)
at akka.actor.UntypedActor.apply(UntypedActor.scala:93)
at akka.actor.ActorCell.invoke(ActorCell.scala:619)
at akka.dispatch.Mailbox.processMailbox(Mailbox.scala:196)
at akka.dispatch.Mailbox.run(Mailbox.scala:178)
at akka.dispatch.ForkJoinExecutorConfigurator$MailboxExecutionTask.exec(AbstractDispatcher.scala:505)
at akka.jsr166y.ForkJoinTask.doExec(ForkJoinTask.java:259)
at akka.jsr166y.ForkJoinPool$WorkQueue.runTask(ForkJoinPool.java:974)
at akka.jsr166y.ForkJoinPool.runWorker(ForkJoinPool.java:1478)
at akka.jsr166y.ForkJoinWorkerThread.run(ForkJoinWorkerThread.java:104)
Akka將序列化演員的狀態。你是否需要應用程序的這部分是有狀態的?如果不是,那麼也許最好使用演員以外的東西。 –
你的意思是我不應該用演員來連接數據庫嗎? – angelokh
如果需要,您絕對可以從演員連接到數據庫。它看起來像你的堆棧跟蹤,它不是序列化,而是觸發lazy屬性,而是'MemberInfo.getType'方法中的一些東西。是對的嗎? –