3

考虑以下代码：

@RestController
@RequestMapping("/timeout")
public class TestController {

    @Autowired
    private TestService service;

    @GetMapping("/max10secs")
    public String max10secs() {
        //In some cases it can take more than 10 seconds
        return service.call();
    }
}

@Service
public class TestService {

    public String call() {
        //some business logic here
        return response;
    }
}

我想要完成的是，如果call来自10 秒的方法TestService需要超过 10 秒，我想取消它并生成带有HttpStatus.REQUEST_TIMEOUT代码的响应。

4

2 回答 2

6

我设法做到了，但我不知道是否有任何概念或实践缺陷是它所遵循的......

一、spring-async的配置

@Configuration
@EnableAsync
public class AsyncConfig implements AsyncConfigurer {

    @Bean(name = "threadPoolTaskExecutor")
    public Executor threadPoolTaskExecutor() {

        ThreadPoolTaskExecutor pool = new ThreadPoolTaskExecutor();
        pool.setCorePoolSize(10);
        pool.setMaxPoolSize(10);
        pool.setWaitForTasksToCompleteOnShutdown(true);
        return pool;
    }

    @Override
    public Executor getAsyncExecutor() {
        return new SimpleAsyncTaskExecutor();
    }
}

接下来，Controller 和 Service 修改：

@RestController
@RequestMapping("/timeout")
public class TestController {

    @Autowired
    private TestService service;

    @GetMapping("/max10secs")
    public String max10secs() throws InterruptedException, ExecutionException {
        Future<String> futureResponse = service.call();
        try {
            //gives 10 seconds to finish the methods execution
            return futureResponse.get(10, TimeUnit.SECONDS);
        } catch (TimeoutException te) {
            //in case it takes longer we cancel the request and check if the method is not done
            if (futureResponse.cancel(true) || !futureResponse.isDone())
                throw new TestTimeoutException();
            else {
                return futureResponse.get();
            }
        }
    }
}

@Service
public class TestService {

    @Async("threadPoolTaskExecutor")
    public Future<String> call() {
        try{
            //some business logic here
            return new AsyncResult<>(response);
        } catch (Exception e) {
            //some cancel/rollback logic when the request is cancelled
            return null;
        }
    }
}

最后生成TestTimeoutException：

@ResponseStatus(value = HttpStatus.REQUEST_TIMEOUT, reason = "too much time")
public class TestTimeoutException extends RuntimeException{ }
于 2018-08-30T13:53:08.413 回答
2

通过DeferredResult还有另一种解决方案。

测试控制器.java

@RestController
@RequestMapping("/timeout")
public class TestController
{

    @Autowired
    private TestService service;


    @GetMapping("/max10secs")
    public DeferredResult<String> max10secs()
    {
        //In some cases it can take more than 10 seconds
        return service.call();
    }
}

测试服务.java

@Service
public class TestService
{

    public DeferredResult<String> call()
    {
        DeferredResult<String> result = new DeferredResult(10000L);
        //some business logic here
        result.onTimeout(()->{
           // do whatever you want there
        });
        result.setResult("test");
        return result;
    }
}

这样，控制器只有在调用result.setResult("test");时才会返回实际结果。.

如您所见，在超时的情况下（超时值在 DeferredResult 对象的构造函数中定义，以毫秒为单位）将执行一个回调，您可以在其中抛出任何异常或返回另一个对象（在您的情况下为HttpStatus.REQUEST_TIMEOUT）。

您可以在此处阅读 Spring 中的 DeferredResult 。

于 2018-08-30T14:28:01.377 回答