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我正在使用 Spring 4.2.3 AsyncRestTemplate.exchange() 调用一些需要几秒钟的 API，并且我期望 listenableFuture.get(1, TimeUnit.SECONDS) 将阻塞 1 秒钟然后抛出 TimeOutException。

相反，listenableFuture.get() 将阻塞整个 api 调用时间（超过 1 秒）

    AsyncRestTemplate restTemplate = new AsyncRestTemplate();

    ListenableFuture<ResponseEntity<String>> listenableFuture = restTemplate.exchange(URL, HttpMethod.GET, null, String.class);
    log.debug("before callback");

    //...add callbacks

    try{
        log.debug("before blocking");
        listenableFuture.get(1, TimeUnit.SECONDS);
    }catch (InterruptedException e) {
        log.error(":GOT InterruptedException");
    } catch (ExecutionException e) {
        log.error(":GOT ExecutionException");
    } catch (TimeoutException e) {
        log.info(":GOT TimeoutException");
    }

    log.info("FINISHED");

输出：

    09:15:21.596  DEBUG [main] org.springframework.web.client.AsyncRestTemplate:78 - Created asynchronous GET request for "http://localhost:4567/oia/wait?seconds=5"
    09:15:21.666  DEBUG [main] org.springframework.web.client.RestTemplate:720 - Setting request Accept header to [text/plain, application/xml, text/xml, application/json, application/*+xml, application/*+json, */*]
    09:15:21.679  DEBUG [main] com.zazma.flow.utils.FutureTest:74 - before callback
    09:15:21.679  DEBUG [main] com.zazma.flow.utils.FutureTest:95 - before blocking
    09:15:26.709  DEBUG [main] org.springframework.web.client.AsyncRestTemplate:576 - Async GET request for "http://localhost:4567/oia/wait?seconds=5" resulted in 200 (OK)
    09:15:26.711  DEBUG [main] org.springframework.web.client.RestTemplate:101 - Reading [java.lang.String] as "text/html;charset=utf-8" using [org.springframework.http.converter.StringHttpMessageConverter@3a44431a]
    09:15:26.717   INFO [main] com.zazma.flow.utils.FutureTest:105 - FINISHED

这是一个示例，当不是由 AsyncRestTemplate 创建时，ListenableFuture.get() 将按预期工作

SimpleAsyncTaskExecutor te = new SimpleAsyncTaskExecutor();
    ListenableFuture<String> lf = te.submitListenable(() -> {
        Thread.sleep(8000);
        return "OK";
    });

    lf.get(1, TimeUnit.SECONDS);
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1 回答 1

1

你的代码是完全正确的。导致问题的是spring框架中的错误。虽然，我没有设法在 Spring 的问题跟踪器中找到它（或者可能没有记录），但您可以通过更新依赖项来修复它。当然，您的代码将适用于 spring-web >= 4.3.2.RELEASE 的版本。

于 2017-02-16T15:34:20.940 回答