Java发送异步post . Spring 最简单的异步发送(非阻塞post)

Java Spring 怎么异步发送post请求?

如果你跟我一样用的Spring 4.0 版本或以上的话,最简单的方式就是使用

AsyncRestTemplate

我用的是fastJson,来post Json 数据

源码如下,URL 是post 链接, 第二个参数是数据, 第三个参数可以不要

    public StringBuffer postHttpJsonDataAsyn(String URL, JSONObject jsonObject, final JSONObject responseJsonMap) throws IOException {
        // 发送异步post , 非阻塞, 发送完无需等待结果返回
        AsyncRestTemplate client = new AsyncRestTemplate();
        ListenableFuture> listenableFuture = client.postForEntity(URL, new HttpEntity(jsonObject), String.class);
        listenableFuture.addCallback(new SuccessCallback>() {
            @Override
            public void onSuccess(ResponseEntity result) {
                System.out.println("("+result.getStatusCode()+ ":"+result.getStatusCode().getReasonPhrase()+ "):"+result.getBody());
            }
        }, new FailureCallback() {
            @Override
            public void onFailure(Throwable ex) {
                System.out.println(ex);
            }
        });
        return new StringBuffer("finish postHttpJsonDataAsyn post");
    }

    @Override
    public StringBuffer postHttpStringDataAsyn(String URL, String postStr, JSONObject results) throws IOException {
        // 发送异步post , 非阻塞, 发送完无需等待结果返回
        AsyncRestTemplate client = new AsyncRestTemplate();
        ListenableFuture> listenableFuture = client.postForEntity(URL, new HttpEntity(postStr), String.class);
        listenableFuture.addCallback(new SuccessCallback>() {
            @Override
            public void onSuccess(ResponseEntity result) {
                System.out.println("("+result.getStatusCode()+ ":"+result.getStatusCode().getReasonPhrase()+ "):"+result.getBody());

            }
        }, new FailureCallback() {
            @Override
            public void onFailure(Throwable ex) {
                ex.printStackTrace();
                System.out.println(ex);
            }
        });
        return new StringBuffer("finish postHttpStringDataAsyn post");
    }

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