如果有错误请求,Java Rest调用响应主体将丢失

时间:2019-11-20 08:55:33

标签: java rest

我使用任何其他客户端在url下方进行访问,并获得带有响应正文的api响应:400错误请求

INPUT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简单客户端不会显示响应主体。.下面是Java代码..它仅给出400个错误的要求。

 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());         
    }       
}

2 个答案:

答案 0 :(得分:0)

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);

使用“尝试捕获”包装。

答案 1 :(得分:0)

对您有用。会起作用

      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:http://SOME.IP:8008/equipment_api/F0-03- 
        8C-C3-D3-CC/832");
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setRequestProperty("content-type", "application/json");
        con.setRequestProperty("X-RequestID", "1234");         
        con.setDoOutput(true);
        OutputStreamWriter writer = new 
        OutputStreamWriter(con.getOutputStream());         
        writer.write(urlParameters);
        writer.flush();         
        String line;

        //BufferedReader reader = new BufferedReader(new 
        InputStreamReader(con.getInputStream()));

        InputStream is = con.getInputStream();
        if(con.getResponseCode() >= 200 && 299 <= con.getResponseCode()) {
            is = con.getInputStream();
        }else {
            is = con.getErrorStream();
        }

        try (BufferedReader br = new BufferedReader(new InputStreamReader(is, 
       "utf-8"))) {
              while ((line = br.readLine()) != null) {                
                System.out.println(line);
            }     
        }

        writer.close();
       // br.close();         
    }catch(Exception ex) {          
        System.out.println("some error :: "+ex.toString());         
    }  
  }