Dart unittest应在异常时失败

Question

我有一个测试：

     Future f = neo4d.nodes.delete(1);
      f.then(((_) {  
      })).catchError((e){
        expect(e.statusCode, equals(409));
      });
      return f;
    });

由于e.statuscode是404而不是409，所以当前的情况就会崩溃。我希望测试失败，但整个测试套件却因为未捕获的异常而停止。我如何捕获异常（并使测试失败）并阻止它破坏所有其他测试？

这是我运行上面的代码时得到的输出：

[2014-03-06 14:44:32.020] DEBUG http: R10: Received data in 9ms with status 404: [{
  "message" : "Cannot find node with id [1] in database.",
  "exception" : "NodeNotFoundException",
  "fullname" : "org.neo4j.server.rest.web.NodeNotFoundException",
  "stacktrace" : [ "org.neo4j.server.rest.web.DatabaseActions.node(DatabaseActions.java:183)", "org.neo4j.server.rest.web.DatabaseActions.deleteNode(DatabaseActions.java:233)", "org.neo4j.server.rest.web.RestfulGraphDatabase.deleteNode(RestfulGraphDatabase.java:279)", "java.lang.reflect.Method.invoke(Method.java:601)", "org.neo4j.server.rest.transactional.TransactionalRequestDispatcher.dispatch(TransactionalRequestDispatcher.java:139)", "org.neo4j.server.rest.security.SecurityFilter.doFilter(SecurityFilter.java:112)", "java.lang.Thread.run(Thread.java:722)" ]
}]
Uncaught Error: Expected: 409
  Actual: 404

Stack Trace: 
#0      SimpleConfiguration.onExpectFailure (package:unittest/src/simple_configuration.dart:141:7)
#1      _ExpectFailureHandler.fail (package:unittest/src/simple_configuration.dart:15:28)
#2      DefaultFailureHandler.failMatch (package:unittest/src/expect.dart:117:9)
#3      expect (package:unittest/src/expect.dart:75:29)
#4      nodes... (file:///Users/oskbor/Projects/neo4dart/test/alltests.dart:202:15)
#5      _invokeErrorHandler (dart:async/async_error.dart:12)
#6      _Future._propagateToListeners. (dart:async/future_impl.dart:469)
#7      _rootRun (dart:async/zone.dart:683)
#8      _RootZone.run (dart:async/zone.dart:823)
#9      _Future._propagateToListeners (dart:async/future_impl.dart:445)
#10     _Future._propagateMultipleListeners (dart:async/future_impl.dart:384)
#11     _Future._propagateToListeners (dart:async/future_impl.dart:411)
#12     _Future._completeError (dart:async/future_impl.dart:315)
#13     _Future._asyncCompleteError. (dart:async/future_impl.dart:367)
#14     _asyncRunCallback (dart:async/schedule_microtask.dart:18)
#15     _createTimer. (dart:async-patch/timer_patch.dart:11)
#16     _Timer._createTimerHandler._handleTimeout (timer_impl.dart:151)
#17     _Timer._createTimerHandler. (timer_impl.dart:166)
#18     _RawReceivePortImpl._handleMessage (dart:isolate-patch/isolate_patch.dart:93)


Unhandled exception:
Expected: 409
  Actual: 404

#0      _rootHandleUncaughtError.. (dart:async/zone.dart:677)
#1      _asyncRunCallback (dart:async/schedule_microtask.dart:18)
#2      _asyncRunCallback (dart:async/schedule_microtask.dart:21)
#3      _createTimer. (dart:async-patch/timer_patch.dart:11)
#4      _Timer._createTimerHandler._handleTimeout (timer_impl.dart:151)
#5      _Timer._createTimerHandler._handleTimeout (timer_impl.dart:159)
#6      _Timer._createTimerHandler._handleTimeout (timer_impl.dart:159)
#7      _Timer._createTimerHandler. (timer_impl.dart:166)
#8      _RawReceivePortImpl._handleMessage (dart:isolate-patch/isolate_patch.dart:93)

问候奥斯卡

Answer 1

异步函数，即返回未来的函数，可以“抛出”两种不同的方式：

它可以同步投掷，甚至不在第一时间返回一个未来：

Future<int> doSomethingAsync() {
  throw new Exception("No! I don't want to even get started!");
}

或者，它可以异步抛出--即返回一个未来，然后异步抛出一个错误，而不是完成：

Future<int> doSomethingAsync() {
  return new Future.error(new Exception("Here's your future, but it'll fail."));
}

neo4d.nodes.delete(1)

  try {
    neo4d.nodes.delete(1).then(((_) { expect(0, 1);  }));
  }
  catch (e) {
    expect(e.statusCode, equals(409));
  }