从异步休息模板spring返回值

Question

我正在使用spring

创建一个异步休息调用

@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

你基本上可以做两件事。

1. 移除ListenableFutureCallback，然后返回ListenableFuture
2. 创建DeferredResult并在ListenableFutureCallback。
中设置该值

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会添加一个ListenableFutureCallback来填充DeferredResult，最终会得到User。

使用DeferredResult

如果您想要更多地控制要返回的内容，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;
}

Answer 2

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

返回你想要的文本

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();
     }
...
}

很抱歉，如果这不是您的要求。