Question

我用Java构造了这个JSON请求。我将使用我的URL和此请求找到一个休息模板。

{  
   "searchRequest":{  
      "header":{  
         "agency":"1111",
         "agent":"2222";
         "from":"0";
         "size":"15"
      },
      "requestParam":{  
         "firstName":"JOHN",
         "lastName":"PAK",
         "dob":"",
         "driverLicense":"",
         "membershipNumber":"",
         "phoneNumbers": "null",
         "addresses":"null"

      }
   }
}

案例1 ：每当我获得成功的响应时，我都会得到我的rest模板在响应变量中给出的JSON。

public @ResponseBody String mpdValidate(@RequestBody String inputRequest, @RequestHeader String url)
            throws JsonParseException, JsonMappingException, IOException, JSONException {
        System.out.println(inputRequest);
        System.out.println(url);

        String response = null;

            if (url == null || url.isEmpty()) {
                url = "myURL";
            }

            try {
                HttpHeaders headers = new HttpHeaders();
                headers.set("X-ApplicationContext",
                        "{\"userId\":\"user\",\"transactionType\":\"realtime\",\"application\":\"app\",\"subSystem\":\"mpd\",\"address\":\"1.0.0.0\",\"correlationId\":\"0f333c\"} ");
                HttpEntity<String> request = new HttpEntity<String>(inputRequest, headers);

                response = restTemplate.postForObject(url, request, String.class);

            } catch (Exception e) {
                response = e.getMessage();
            }

        return response;
    }

案例2 ：当错误的请求框架且响应失败时，其余模板将返回此响应。

{
   "httpCode": 400,
   "httpMessage": "Bad Request",
   "moreInformation": "Request parameter is null",
   "timeStamp": 1539072063795
}

但是响应变量返回null并进入catch块并抛出null指针异常。

我希望将上述JSON格式的字符串格式添加到我的响应变量中。

有人可以帮忙吗？

Answer 1

try {
    // ...
} catch (HttpClientErrorException expection) {
    response = expection.getResponseBodyAsString();
}

您需要处理HttpClientErrorException（或其父RestClientResponseException）并通过HttpClientErrorException#getResponseBodyAsString提取响应。

Answer 2

您应该尝试这样，

// REST Request
        try {
            restTemplate.postForObject(requestUrl, postBody, Void.class);
        } catch (RestException restException) {
            Logger.error(this, "RestException: " + restException.getRestError().toString());
    response = restException.getRestError().toString();
        }

在字符串变量中从Rest服务获取异常

2 个答案: