如何测试DeferredResult timeoutResult

Question

从一段时间以前,我正在按照Spring博客实施长轮询.

这里我的转换方法具有与以前相同的响应签名,但它现在使用长轮询而不是立即响应:

private Map<String, DeferredResult<ResponseEntity<?>>> requests = new ConcurrentHashMap<>();

@RequestMapping(value = "/{uuid}", method = RequestMethod.GET)
public DeferredResult<ResponseEntity<?>> poll(@PathVariable("uuid") final String uuid) {
    // Create & store a new instance
    ResponseEntity<?> pendingOnTimeout = ResponseEntity.accepted().build();
    DeferredResult<ResponseEntity<?>> deferredResult = new DeferredResult<>(TWENTYFIVE_SECONDS, pendingOnTimeout);
    requests.put(uuid, deferredResult);

    // Clean up poll requests when done
    deferredResult.onCompletion(() -> {
        requests.remove(deferredResult);
    });

    // Set result if already available
    Task task = taskHolder.retrieve(uuid);
    if (task == null)
        deferredResult.setResult(ResponseEntity.status(HttpStatus.GONE).build());
    else
        // Done (or canceled): Redirect to retrieve file contents
        if (task.getFutureFile().isDone())
            deferredResult.setResult(ResponseEntity.created(RetrieveController.uri(uuid)).build());

    // Return result
    return deferredResult;
}

特别是pendingOnTimeout当请求花费太长时间(我之前立即返回)时,我想返回响应,以防止代理切断请求.

现在我觉得我已经按原样运行,但是我想编写一个证明这一点的单元测试.但是,我使用MockMvc(通过webAppContextSetup)的所有尝试都无法为我提供断言我得到accepted标题的方法.例如,当我尝试以下内容时:

@Test
public void pollPending() throws Exception {
    MvcResult result = mockMvc.perform(get("/poll/{uuid}", uuidPending)).andReturn();
    mockMvc.perform(asyncDispatch(result))
            .andExpect(status().isAccepted());
}

我得到以下stacktrace:

java.lang.IllegalStateException:处理程序[public org.springframework.web.context.request.async.DeferredResult> nl.bioprodict.blast.api.PollController.poll(java.lang.String)]的异步结果未在org.springframework.test.web上org.springframework.test.web.servlet.DefaultMvcResult.getAsyncResult(DefaultMvcResult.java:143)的org.springframework.util.Assert.state(Assert.java:392)中指定的timeToWait = 25000 org.springframework.test.web.servlet.request.MockMvcRequestBuilders.asyncDispatch(MockMvcRequestBuilders.java:235)中的.servlet.DefaultMvcResult.getAsyncResult(DefaultMvcResult.java:120)at nl.bioprodict.blast.docs.PollControllerDocumentation.pollPending(PollControllerDocumentation) .java:53)...

与此相关的Spring框架测试,我发现它似乎都使用模拟它似乎:https://github.com/spring-projects/spring-framework/blob/master/spring-web/src/test/java/org/springframework /web/context/request/async/WebAsyncManagerTimeoutTests.java

如何测试DeferredResult timeoutResult的正确处理？

Answer 1

在我的情况下,在通过spring源代码并设置超时(10000毫秒)并获得异步结果后,我就解决了它,as;

 mvcResult.getRequest().getAsyncContext().setTimeout(10000);
 mvcResult.getAsyncResult();

我的整个测试代码是;

MvcResult mvcResult = this.mockMvc.perform(
                                post("<SOME_RELATIVE_URL>")
                                .contentType(MediaType.APPLICATION_JSON)
                                .content(<JSON_DATA>))
                        ***.andExpect(request().asyncStarted())***
                            .andReturn();

***mvcResult.getRequest().getAsyncContext().setTimeout(10000);***
***mvcResult.getAsyncResult();***

this.mockMvc
    .perform(asyncDispatch(mvcResult))
    .andDo(print())
    .andExpect(status().isOk());

希望能帮助到你..

Answer 2

我使用 Spring 4.3遇到了这个问题，并设法找到了一种方法来从单元测试中触发超时回调。在获取MvcResult,之后，在调用 之前asyncDispatch()，您可以插入如下代码：

MockAsyncContext ctx = (MockAsyncContext) mvcResult.getRequest().getAsyncContext();
for (AsyncListener listener : ctx.getListeners()) {
    listener.onTimeout(null);
}

请求的异步侦听器之一将调用DeferredResult的超时回调。

所以你的单元测试看起来像这样：

@Test
public void pollPending() throws Exception {
    MvcResult result = mockMvc.perform(get("/poll/{uuid}", uuidPending)).andReturn();
    MockAsyncContext ctx = (MockAsyncContext) result.getRequest().getAsyncContext();
    for (AsyncListener listener : ctx.getListeners()) {
        listener.onTimeout(null);
    }
    mockMvc.perform(asyncDispatch(result))
            .andExpect(status().isAccepted());
}