HttpPost请求将json作为请求体传入的简单处理方法

 1 public static String sendHttpPost(String url, String body) throws Exception { 2 CloseableHttpClient httpClient = HttpClients.createDefault(); 3 HttpPost httpPost = new HttpPost(url); 4 httpPost.addHeader("Content-Type", "application/json"); 5 httpPost.setEntity(new StringEntity(body)); 6 ?7 CloseableHttpResponse response = httpClient.execute(httpPost); 8 System.out.println(response.getStatusLine().getStatusCode() + "\n"); 9 HttpEntity entity = response.getEntity();10 String responseContent = EntityUtils.toString(entity, "UTF-8"); 11 System.out.println(responseContent);12 13 response.close();14 httpClient.close();15 return responseContent;16 }

 1 ????public static void main(String[] args) { 2 ????????//测试公司的API接口，将json当做一个字符串传入httppost的请求体 3 ????????String result = null; 4 ????????HttpClient client = HttpClients.createDefault(); 5 ????????URIBuilder builder = new URIBuilder(); 6 ????????URI uri = null; 7 ????????try { 8 ????????????uri = builder.setScheme("http") 9 ??????????????????????.setHost("xxx.xxx.xxx.xxx:xxxx")10 ??????????????????????.setPath("/api/authorize/login")11 ??????????????????????.build();12 ????????????13 ????????????HttpPost post = new HttpPost(uri);14 ????????????//设置请求头15 ????????????post.setHeader("Content-Type", "application/json");16 ????????????String body = "{\"Key\": \"\",\"Secret\": \"\"}";17 ????????????//设置请求体18 ????????????post.setEntity(new StringEntity(body));19 ????????????//获取返回信息20 ????????????HttpResponse response = client.execute(post);21 ????????????result = response.toString();22 ????????} catch (Exception e) {23 ????????????System.out.println("接口请求失败"+e.getStackTrace());24 ????????}25 ????????System.out.println(result);26 ????}