如果请求错误，java 休息调用响应主体会丢失

Question

我使用任何休息客户端点击下面的网址，我得到了 api 响应:400 带响应正文的错误请求

输入参数

POST http://SOME.IP:8008/equipment_api/F0-03-8C-C3-D3-CC/832
HEADERS
Content-Type application/json
X-RequestID 1234

BODY
{
"items":[{"updateValue":1, "updateKey": "RESETDEV"}],
"sync":"false"
}

Response :400 Bad Request
{
"error": "RESETDEV is not a valid key."
}

但是java简单客户端不显示响应正文..下面是java代码..它只是给出400错误请求。

 public static void main(String[] args) {
    try {           
        String urlParameters = "  {\r\n" + 
                "\"items\":[{\"updateValue\":\"Hi\", \"updateKey\": \"RESETDEV\"}],\r\n" + 
                "\"sync\":false\r\n" + 
                "}";
        URL url = new URL("http://SOME.IP:8008/equipment_api/F0-03-8C-C3-D3-CC/832");
        URLConnection conn = url.openConnection();
        conn.setRequestProperty("content-type", "application/json");
        conn.setRequestProperty("X-RequestID", "1234");         
        conn.setDoOutput(true);
        OutputStreamWriter writer = new OutputStreamWriter(conn.getOutputStream());         
        writer.write(urlParameters);
        writer.flush();         
        String line;
        BufferedReader reader = new BufferedReader(new InputStreamReader(conn.getInputStream()));

        while ((line = reader.readLine()) != null) {                
            System.out.println(line);
        }           
        writer.close();
        reader.close();         
    }catch(Exception ex) {          
        System.out.println("some error :: "+ex.toString());         
    }       
}

Answer 1

尝试使用以下代码

final String uri = "http://SOME.IP:8008/equipment_api/F0-03-8C-C3-D3-CC/832";

RestTemplate restTemplate = new RestTemplate();

HttpHeaders headers = new HttpHeaders(); 
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON)); 
HttpEntity<String> entity = new HttpEntity<String>("X-RequestID", "1234");

ResponseEntity<String> result = restTemplate.exchange(uri, HttpMethod.GET, entity, String.class);
System.out.println(result);

用 Try Catch 包裹起来。