嘲笑vertx.io异步处理程序

Question

当我同步时，我编写了单元测试嘲笑持久性部分并检查呼叫者的行为。这是我通常做什么的示例：

@Mock
private OfferPersistenceServiceImpl persistenceService;
@Inject
@InjectMocks
private OfferServiceImpl offerService;
...
@Test
public void createInvalidOffer() {
  offer = new Offer(null, null, null, null, null, 4, 200D, 90D);
  String expectedMessage = Offer.class.getName() + " is not valid: " + offer.toString();
  Mockito.when(persistenceService.create(offer)).thenThrow(new IllegalArgumentException(expectedMessage));
  Response response = offerService.create(offer);
  Mockito.verify(persistenceService, Mockito.times(1)).create(offer);
  Assert.assertEquals(INVALID_INPUT, response.getStatus());
  String actualMessage = response.getEntity().toString();
  Assert.assertEquals(expectedMessage, actualMessage);
}

，但是现在我爱上了vertx.io（我很新），我想成为异步。好的。但是vertx有处理程序，因此要模拟的新持久性组件如下：

...
mongoClient.insert(COLLECTION, offer, h-> {
  ...
});

所以我猜想如何模拟处理程序h来测试使用该mongoClient的类，甚至是使用vertx.io测试的正确方法。我正在使用vertx.io 3.5.0，junit 4.12和mockito 2.13.0。谢谢。

更新我试图遵循tsegimond的建议，但我无法了解Mockito的Answer和ArgumentCaptor如何帮助我。这是我到目前为止尝试的。使用ArgumentCaptor：

JsonObject offer = Mockito.mock(JsonObject.class);
Mockito.when(msg.body()).thenReturn(offer);         
Mockito.doNothing().when(offerMongo).validate(offer);
RuntimeException rex = new RuntimeException("some message");
...
ArgumentCaptor<Handler<AsyncResult<String>>> handlerCaptor =
ArgumentCaptor.forClass(Handler.class);
ArgumentCaptor<AsyncResult<String>> asyncResultCaptor =
ArgumentCaptor.forClass(AsyncResult.class);
offerMongo.create(msg);
Mockito.verify(mongoClient,
Mockito.times(1)).insert(Mockito.anyString(), Mockito.any(), handlerCaptor.capture());
Mockito.verify(handlerCaptor.getValue(),
Mockito.times(1)).handle(asyncResultCaptor.capture());
Mockito.when(asyncResultCaptor.getValue().succeeded()).thenReturn(false);
Mockito.when(asyncResultCaptor.getValue().cause()).thenReturn(rex);
Assert.assertEquals(Json.encode(rex), msg.body().encode());

使用Answer：

ArgumentCaptor<AsyncResult<String>> handlerCaptor =
ArgumentCaptor.forClass(AsyncResult.class);
AsyncResult<String> result = Mockito.mock(AsyncResult.class);
Mockito.when(result.succeeded()).thenReturn(true);
Mockito.when(result.cause()).thenReturn(rex);
Mockito.doAnswer(new Answer<MongoClient>() {
  @Override
  public MongoClient answer(InvocationOnMock invocation) throws Throwable {
    ((Handler<AsyncResult<String>>)
    invocation.getArguments()[2]).handle(handlerCaptor.capture());
        return null;
      }
    }).when(mongoClient).insert(Mockito.anyString(), Mockito.any(),
Mockito.any());
userMongo.create(msg);
Assert.assertEquals(Json.encode(rex), msg.body().encode());

现在我感到困惑。有没有办法模拟AsyncResult让它在succeed()上返回false？

Answer 1

最后，我有几次进行调查，然后做了。这是我的解决方案。

@RunWith(PowerMockRunner.class)
@PowerMockRunnerDelegate(VertxUnitRunner.class)
@PrepareForTest({ MongoClient.class })
public class PersistenceTest {
private MongoClient mongo;
private Vertx vertx;
@Before
public void initSingleTest(TestContext ctx) throws Exception {
  vertx = Vertx.vertx();
  mongo = Mockito.mock(MongoClient.class);
  PowerMockito.mockStatic(MongoClient.class);
  PowerMockito.when(MongoClient.createShared(Mockito.any(), Mockito.any())).thenReturn(mongo);
  vertx.deployVerticle(Persistence.class, new DeploymentOptions(), ctx.asyncAssertSuccess());
}
@SuppressWarnings("unchecked")
@Test
public void loadSomeDocs(TestContext ctx) {
  Doc expected = new Doc();
  expected.setName("report");
  expected.setPreview("loremipsum");
  Message<JsonObject> msg = Mockito.mock(Message.class);
  Mockito.when(msg.body()).thenReturn(JsonObject.mapFrom(expected));
  JsonObject result = new JsonObject().put("name", "report").put("preview", "loremipsum");
  AsyncResult<JsonObject> asyncResult = Mockito.mock(AsyncResult.class);
  Mockito.when(asyncResult.succeeded()).thenReturn(true);
  Mockito.when(asyncResult.result()).thenReturn(result);
  Mockito.doAnswer(new Answer<AsyncResult<JsonObject>>() {
    @Override
    public AsyncResult<JsonObject> answer(InvocationOnMock arg0) throws Throwable {
    ((Handler<AsyncResult<JsonObject>>) arg0.getArgument(3)).handle(asyncResult);
    return null;
    }
  }).when(mongo).findOne(Mockito.any(), Mockito.any(), Mockito.any(), Mockito.any());
  Async async = ctx.async();
  vertx.eventBus().send("persistence", new JsonObject(), msgh -> {
    if (msgh.failed()) {
    System.out.println(msgh.cause().getMessage());
    }
    ctx.assertTrue(msgh.succeeded());
    ctx.assertEquals(expected, Json.decodeValue(msgh.result().body().toString(), Doc.class));
    async.complete();
  });
  async.await();
  }
}

使用Powemockito模拟MongoClient.createShared静态方法，因此在开始时，您将获得模拟。嘲笑异步处理程序是要编写的一些代码。如您所见，嘲笑从Message<JsonObject> msg = Mockito.mock(Message.class);开始，并以Mockito.doAnswer(new Answer...结束。在Answer的方法中，选择处理程序参数并强迫它处理您的异步结果，然后您完成了。

Answer 2

通常，我会使用评论来发布此信息，但是格式丢失了。可接受的解决方案非常有效，只需注意，可以使用Java 8 来简化一点，您可以使用实际对象而不是JSON。

doAnswer((Answer<AsyncResult<List<Sample>>>) arguments -> {
            ((Handler<AsyncResult<List<Sample>>>) arguments.getArgument(1)).handle(asyncResult);
            return null;
        }).when(sampleService).findSamplesBySampleFilter(any(), any());

getArgument（1），在诸如：

之类的方法中指的是处理程序参数的索引

@Fluent
@Nonnull
SampleService findSamplesBySampleFilter(@Nonnull final SampleFilter sampleFilter,
                                  @Nonnull final Handler<AsyncResult<List<Sample>>> resultHandler);