Question

我在Scala / Play中使用Cassandra作为数据库的Web应用程序。我在测试潜在错误时遇到了以下问题。

以下是我项目的组成部分

Application.scala - ＆gt;基本控制器
Model.scala - ＆gt;持有业务逻辑
CassandraClient.scala - ＆gt;连接的逻辑 cassandra并在cassandra上运行查询

CassandraClient.scala如下所示

object CassandraClient {

private val cluster = Cluster.builder()
    .withLoadBalancingPolicy(
      new WhiteListPolicy(new RoundRobinPolicy(), nodes))
    .withSocketOptions(socketOptions)
    .addContactPointsWithPorts(nodes)
    .withCredentials(CASSANDRA_USERNAME, CASSANDRA_PWD).build()
}

 private val session = cluster.connect()

def getValueFromCassandraTable(token:String) = {
 var query = QueryBuilder.select.all()...
 seesion.execute(query)
}

当我第一次调用getValueFromCassandraTable时，首先建立了Cassandra连接。由于CassandraClient是一个Object，连接到Cassandra的逻辑只被调用一次。

Model.Scala有一些代码来处理session.execute返回的未来。

例如：在Model.scala中查看下面的示例代码。

    def getTitles(titles: String)(implicit ctxt: ExecutionContext): Future[List[<Sometype>]] = {
        try{
            CassandraClient.getValueFromCassandraTable(token).toScalaFuture.map { rows =>
                  rows.map(row => row("value").toList
                }.recover { case e: Exception =>List()}
              }
        } catch{case e:Exception => 
           Logger.info("THIS DOES NOT EXECUTE??")
           Future{List()}}
    }

现在，当第一次执行上面的示例代码时，session.execute（）会被执行。然后执行getValueFromCassandraTable。所以我认为执行的流程是 Models.scala上面的代码片段 - ＆gt; session.execute（） - ＆gt; getValueFromCassandraTable（）

因此，如果session.execute失败，我应该能够在try catch异常块中捕获它。但令我惊讶的是，当session.execute失败时，不会执行catch块。相反，play框架会引发异常。有人可以解释这种行为。

异常堆栈

Caused by: java.lang.RuntimeException: java.lang.ExceptionInInitializerError
    at play.api.mvc.ActionBuilder$$anon$1.apply(Action.scala:498) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$$anonfun$apply$1$$anonfun$apply$4$$anonfun$apply$5.apply(Action.scala:105) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$$anonfun$apply$1$$anonfun$apply$4$$anonfun$apply$5.apply(Action.scala:105) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.utils.Threads$.withContextClassLoader(Threads.scala:21) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$$anonfun$apply$1$$anonfun$apply$4.apply(Action.scala:104) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$$anonfun$apply$1$$anonfun$apply$4.apply(Action.scala:103) ~[play_2.10-2.4.2.jar:2.4.2]
    at scala.Option.map(Option.scala:145) ~[scala-library-2.10.5.jar:na]
    at play.api.mvc.Action$$anonfun$apply$1.apply(Action.scala:103) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$$anonfun$apply$1.apply(Action.scala:96) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.libs.iteratee.DoneIteratee$$anonfun$mapM$2.apply(Iteratee.scala:741) ~[play-iteratees_2.10-2.4.2.jar:2.4.2]
Caused by: java.lang.ExceptionInInitializerError: null
    at models.Model$.getTitles(Model.scala:121) ~[classes/:na]
    at models.Model$.getMatchingTitles(Model.scala:48) ~[classes/:na]
    at models.Model$.getMatchingTitles(Model.scala:56) ~[classes/:na]
    at controllers.Application$$anonfun$searchTitle$1.apply(Application.scala:15) ~[classes/:na]
    at controllers.Application$$anonfun$searchTitle$1.apply(Application.scala:15) ~[classes/:na]
    at play.api.mvc.ActionBuilder$$anonfun$async$1.apply(Action.scala:456) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.ActionBuilder$$anonfun$async$1.apply(Action.scala:456) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$.invokeBlock(Action.scala:533) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.Action$.invokeBlock(Action.scala:530) ~[play_2.10-2.4.2.jar:2.4.2]
    at play.api.mvc.ActionBuilder$$anon$1.apply(Action.scala:493) ~[play_2.10-2.4.2.jar:2.4.2]
Caused by: com.datastax.driver.core.exceptions.AuthenticationException: Authentication error on host /10.65.5.44:9042: Username and/or password are incorrect
    at com.datastax.driver.core.Connection$8.apply(Connection.java:368) ~[cassandra-driver-core-2.1.6.jar:na]
    at com.datastax.driver.core.Connection$8.apply(Connection.java:338) ~[cassandra-driver-core-2.1.6.jar:na]
    at com.google.common.util.concurrent.Futures$ChainingListenableFuture.run(Futures.java:861) ~[guava-16.0.1.jar:na]
    at com.google.common.util.concurrent.MoreExecutors$SameThreadExecutorService.execute(MoreExecutors.java:297) ~[guava-16.0.1.jar:na]
    at com.google.common.util.concurrent.ExecutionList.executeListener(ExecutionList.java:156) ~[guava-16.0.1.jar:na]
    at com.google.common.util.concurrent.ExecutionList.execute(ExecutionList.java:145) ~[guava-16.0.1.jar:na]
    at com.google.common.util.concurrent.AbstractFuture.set(AbstractFuture.java:185) ~[guava-16.0.1.jar:na]
    at com.datastax.driver.core.Connection$Future.onSet(Connection.java:1170) ~[cassandra-driver-core-2.1.6.jar:na]
    at com.datastax.driver.core.Connection$Dispatcher.channelRead0(Connection.java:1000) ~[cassandra-driver-core-2.1.6.jar:na]
    at com.datastax.driver.core.Connection$Dispatcher.channelRead0(Connection.java:922) ~[cassandra-driver-core-2.1.6.jar:na]

Modle.scala中的Lin＃121是一个调用CassandrClient.getValueFromCassandraTable的人。我通过传递连接到cassandra的错误密码来模仿异常，因此获得AuthroizationException。我希望在Try catch中捕获此异常。但它没有被抓住。

（请注意，这与Future无关，因为当异常发生时甚至没有调用session.executAysnc。所以期货尚未发挥作用。）

- 编辑 -

看起来Play正在吞噬Exception并抛出一个Error对象。不知道为什么会这样。

Answer 1

session.execute（）可能会因IOError之类的错误而失败。你只能抓住异常。

检查您是否没有错误，只是琐碎的异常。如果你要在你的问题中添加堆栈跟踪，那就太好了。

您可以尝试捕获任何Throwable（不要在实际项目中执行，您只能在调试时尝试）

try {
  .... your code ...
} catch {
  case _ => errorHandler(e)
}

只有在未来的情况下它才会捕获错误。

修改

我的cassandra数据库上的实例。它在连接到cassandra期间出现异常时显示用户列表或“出错”字符串。例如，错误的节点或cassandra正在关闭。

package controllers import com.datastax.driver.core.Cluster import com.datastax.driver.core.Session import scala.collection.JavaConversions._ import play.api._ import play.api.mvc._ class Application extends Controller { def index = Action { try{ var b = new StringBuilder for (row <- CassandraClient.getValueFromCassandraTable()) { b ++= row.getString("user_id") b ++= "\n" } Ok(b.toString()) } catch { case _ => InternalServerError("something going wrong") } } } object CassandraClient { private val cluster = Cluster.builder() .addContactPoint("localhost") .withPort(9042) .build() val session = cluster.connect() def getValueFromCassandraTable() = { session.execute("SELECT * FROM mykeyspace.users") } }

Answer 2

也许你有InterruptedException

例如代码不打印＆＃34; InterruptedException test＆＃34;如果你不使用Await。

 try {
   val exception = Future {
     throw new InterruptedException("exception")
   }.recover { case e: Exception => "ok" }

  exception.onComplete {
    case Success(a) => println(a)
    case Failure(err) => println(err)
  }
  //Await.result(exception, 1 seconds) 

 } catch {
   case e: Exception => println("InterruptedException test")
 }

了解play / scala错误处理

2 个答案: