我正在按照不久前的 Spring 博客实现长轮询。
这里我转换后的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,当请求花费太长时间时,我想返回
pendingOnTimeout
响应(我之前立即返回过),以防止代理切断请求。
现在我认为我已经按原样工作了,但我想编写一个单元测试来证实这一点。然而,我使用 MockMvc (通过 webAppContextSetup)的所有尝试都未能为我提供一种断言我获得
accepted
标头的方法。例如,当我尝试以下操作时:
@Test
public void pollPending() throws Exception {
MvcResult result = mockMvc.perform(get("/poll/{uuid}", uuidPending)).andReturn();
mockMvc.perform(asyncDispatch(result))
.andExpect(status().isAccepted());
}
我得到以下堆栈跟踪:
java.lang.IllegalStateException:处理程序的异步结果 [public org.springframework.web.context.request.async.DeferredResult> nl.bioprodict.blast.api.PollController.poll(java.lang.String)] 期间未设置指定的 timeToWait=25000 在 org.springframework.util.Assert.state(Assert.java:392) 在 org.springframework.test.web.servlet.DefaultMvcResult.getAsyncResult(DefaultMvcResult.java:143) 在 org.springframework.test.web.servlet.DefaultMvcResult.getAsyncResult(DefaultMvcResult.java:120) 在 org.springframework.test.web.servlet.request.MockMvcRequestBuilders.asyncDispatch(MockMvcRequestBuilders.java:235) 在nl.bioprodict.blast.docs.PollControllerDocumentation.pollPending(PollControllerDocumentation.java:53) ...
与此相关的 Spring 框架测试似乎都使用了模拟: 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?
就我而言,在查看 Spring 源代码并设置超时(10000 毫秒)并获取异步结果后,为我解决了这个问题,如下;
mvcResult.getRequest().getAsyncContext().setTimeout(10000);
mvcResult.getAsyncResult();
// My whole test code was;
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());
希望有帮助..
我使用Spring 4.3遇到了这个问题,并设法找到了一种在单元测试中触发超时回调的方法。获取
MvcResult
后,在调用 asyncDispatch()
之前,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。
所以你的单元测试将如下所示:
@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());
}