Return 来自异步 rest 模板的值 spring

Question

我正在使用 spring

创建异步休息调用

@GetMapping(path = "/testingAsync")
public String value() throws ExecutionException, InterruptedException, TimeoutException {
    AsyncRestTemplate restTemplate = new AsyncRestTemplate();
    String baseUrl = "https://api.github.com/users/XXX";
    HttpHeaders requestHeaders = new HttpHeaders();
    requestHeaders.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    String value = "";

    HttpEntity entity = new HttpEntity("parameters", requestHeaders);
    ListenableFuture<ResponseEntity<User>> futureEntity = restTemplate.getForEntity(baseUrl, User.class);

    futureEntity.addCallback(new ListenableFutureCallback<ResponseEntity<User>>() {
        @Override
        public void onSuccess(ResponseEntity<User> result) {
            System.out.println(result.getBody().getName());
            // instead of this how can i return the value to the user ?
        }

        @Override
        public void onFailure(Throwable ex) {

        }
    });

    return "DONE"; // instead of done i want to return value to the user comming from the rest call 
}

有什么方法可以将 ListenableFuture 转换为使用 java 8 中使用的 CompletableFuture？

Answer 1

我不太了解 Spring 中的异步调用，但我想你可以 return 通过 ResponseBody

看起来像这样：

@GetMapping(path = "/testingAsync")
@ResponseBody
public String value() throws ExecutionException, InterruptedException, TimeoutException {
...
...
     @Override
     public void onSuccess(ResponseEntity<User> result) {
         return result.getBody().getName();
     }
...
}

抱歉，如果这不是您要问的问题。

Answer 2

您基本上可以做两件事。

1. 删除 ListenableFutureCallback 并简单地 return ListenableFuture
2. 创建一个 DeferredResult 并在 ListenableFutureCallback 中设置它的值。

返回 ListenableFuture

@GetMapping(path = "/testingAsync")
public ListenableFuture<ResponseEntity<User>> value() throws ExecutionException, InterruptedException, TimeoutException {
    AsyncRestTemplate restTemplate = new AsyncRestTemplate();
    String baseUrl = "https://api.github.com/users/XXX";
    HttpHeaders requestHeaders = new HttpHeaders();
    requestHeaders.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    String value = "";

    HttpEntity entity = new HttpEntity("parameters", requestHeaders);
    return restTemplate.getForEntity(baseUrl, User.class);
}

Spring MVC 将添加一个 ListenableFutureCallback 本身来填充 DeferredResult，最终您将得到一个 User。

使用 DeferredResult

如果您想更好地控制要 return 的内容，您可以使用 DeferredResult 并自行设置值。

@GetMapping(path = "/testingAsync")
public DeferredResult<String> value() throws ExecutionException, InterruptedException, TimeoutException {
    AsyncRestTemplate restTemplate = new AsyncRestTemplate();
    String baseUrl = "https://api.github.com/users/XXX";
    HttpHeaders requestHeaders = new HttpHeaders();
    requestHeaders.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
    String value = "";

    HttpEntity entity = new HttpEntity("parameters", requestHeaders);
    final DeferredResult<String> result = new DeferredResult<>();
    ListenableFuture<ResponseEntity<User>> futureEntity = restTemplate.getForEntity(baseUrl, User.class);

    futureEntity.addCallback(new ListenableFutureCallback<ResponseEntity<User>>() {
        @Override
        public void onSuccess(ResponseEntity<User> result) {
            System.out.println(result.getBody().getName());
            result.setResult(result.getBody().getName());
        }

        @Override
        public void onFailure(Throwable ex) {
            result.setErrorResult(ex.getMessage());
        }
    });

    return result;
}