我有这个测试:
Future f = neo4d.nodes.delete(1); f.then(((_) { })).catchError((e){ expect(e.statusCode, equals(409)); }); return f; });
目前正在爆炸,因为e.statusCode
是404而不是409.我希望测试失败但是由于未捕获的异常而导致整个测试套件停止。
我如何捕获异常(并且未通过测试)并阻止其爆炸所有其他测试?
这是我运行上面代码时得到的输出:
[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)
关于奥斯卡
答案 0 :(得分:3)
异步函数,即返回Future的函数,可以“抛出”两种不同的方式:
它可以同步抛出,甚至不会首先返回Future:
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Future,然后异步抛出错误而不是完成:
Future<int> doSomethingAsync() {
return new Future.error(new Exception("Here's your future, but it'll fail."));
}
您的异步电话
neo4d.nodes.delete(1)
必须是前一种类型:它甚至在没有返回Future的情况下立即抛出。这就是为什么异常不会被.catchError
捕获,而是炸毁整个测试套件。
您想要更新neo4d.nodes.delete
并使其异步抛出。或者,如果无法做到这一点,请将测试包装在一个良好的旧同步try-catch中:
try {
neo4d.nodes.delete(1).then(((_) { expect(0, 1); }));
}
catch (e) {
expect(e.statusCode, equals(409));
}