Question

是否有一种模拟易于休息的异步HTTP请求的正式方法？

示例代码：

@GET
@Path("test")
public void test(@Suspended final AsyncResponse response) {

    Thread t = new Thread() {
        @Override
        public void run()
        {

            try {
                Response jaxrs = Response.ok("basic").type(MediaType.APPLICATION_JSON).build();
                response.resume(jaxrs);
            }
            catch (Exception e)
            {
                e.printStackTrace();
            }
        }

    };
    t.start();
}

我以这种方式拒绝模拟休息 - 容易的请求：

@Test
public void test() throws Exception {
    /**
     * mock
     */
    Dispatcher dispatcher = MockDispatcherFactory.createDispatcher();

    dispatcher.getRegistry().addSingletonResource(action);
    MockHttpRequest request = MockHttpRequest.get("/hello/test");
    request.addFormHeader("X-FORWARDED-FOR", "122.122.122.122");
    MockHttpResponse response = new MockHttpResponse();

    /**
     * call
     */
    dispatcher.invoke(request, response);

    /**
     * verify
     */
    System.out.println("receive content:"+response.getContentAsString());
}

但是它没有用。我在单元测试期间遇到了BadRequestException。

模拟易于休息的异步HTTP请求的正确方法是什么？

Answer 1

By reading rest-easy's source code, I finally find a way to work around with asynchronous HTTP request :

@Test
public void test() throws Exception {
    /**
     * mock
     */
    Dispatcher dispatcher = MockDispatcherFactory.createDispatcher();

    dispatcher.getRegistry().addSingletonResource(action);
    MockHttpRequest request = MockHttpRequest.get("/hello/test");
    request.addFormHeader("X-FORWARDED-FOR", "122.122.122.122");
    MockHttpResponse response = new MockHttpResponse();
    // Add following two lines !!!
    SynchronousExecutionContext synchronousExecutionContext = new SynchronousExecutionContext((SynchronousDispatcher)dispatcher, request, response );
    request.setAsynchronousContext(synchronousExecutionContext);

    /**
     * call
     */
    dispatcher.invoke(request, response);

    /**
     * verify
     */
    System.out.println("receive content:"+response.getContentAsString());
}

The output of response is correct !!!

如何模拟rest-easy的异步HTTP请求？

1 个答案: