java – 无法编写JSON:没有为类org.json.JSONObject找到序列化
发布时间:2020-12-15 04:40:23 所属栏目:Java 来源:网络整理
导读:我已经将响应设置为 JSON但是得到了这个 Could not write JSON: No serializer found for class org.json.JSONObject and no properties discovered to create BeanSerializer (to avoid exception,disable SerializationFeature.FAIL_ON_EMPTY_BEANS) @Requ
我已经将响应设置为
JSON但是得到了这个
@RequestMapping(value = "/customerlist",method = RequestMethod.POST) public ResponseGenerator getCustomerList() { ResponseGenerator responseGenerator = new ResponseGenerator(); try { responseGenerator.setCode(StatusCode.SUCCESS.code); responseGenerator.setMessage(StatusCode.SUCCESS.message); responseGenerator.setStatus(ResponseStatus.SUCCESS); JSONObject data = userService.getUserList(); responseGenerator.setJSONData(data); return responseGenerator; //error here } catch (Exception e) { logger.error("Error while getting Customer List : ",e); e.printStackTrace(); responseGenerator.setCode(StatusCode.GENERAL_ERROR.code); responseGenerator.setMessage(StatusCode.GENERAL_ERROR.message); responseGenerator.setStatus(ResponseStatus.FAIL); return responseGenerator; } } userService.getUserList(): public JSONObject jsonResp; public JSONObject getUserList() throws Exception{ jsonResp =new JSONObject(); //List<JSONObject> customers = new ArrayList<JSONObject>(); JSONObject jsonResponse = erpNextAPIClientService.getCustomerList(); //ObjectMapper objectMapper = new ObjectMapper(); //objectMapper.setVisibility(PropertyAccessor.FIELD,Visibility.ANY); //JSONArray jsonArray = objectMapper.convertValue(jsonResponse.get("data"),JSONArray.class); JSONArray jsonArray = jsonResponse.getJSONArray("data"); //JSONArray jsonArray =new Gson().fromJson(jsonResponse.get("data").toString(),JSONArray.class); for (int i = 0; i < jsonArray.length(); i++) { JSONObject cust = erpNextAPIClientService.getUser(jsonArray.getJSONObject(i).get("name").toString()); JSONObject custAddress =erpNextAPIClientService.getCustomerAddress(jsonArray.getJSONObject(i).get("name").toString()); JSONObject custData = new JSONObject(cust.getString("data")); JSONObject custAddressData = new JSONObject(custAddress.getString("data")); custData.accumulate("bill_to_address_line_one",custAddressData.get("address_line1")); custData.accumulate("bill_to_address_line_two",custAddressData.get("address_line2")); custData.accumulate("bill_to_city",custAddressData.get("city")); custData.accumulate("bill_to_state",custAddressData.get("state")); custData.accumulate("bill_to_zip",custAddressData.get("pincode")); custData.accumulate("bill_to_country",custAddressData.get("country")); jsonResp.put("data",custData); System.out.println(custData.toString()); //customers.add(custData); } return jsonResp; } 解决方法
这将抛出一个错误,因为JSONObject不公开默认的getter.
虽然可以采取一种解决方法来避免这种情况. 您需要更改ResponseGenerator类以接受Map< String,Object>而不是JSONObject. responseGenerator.setJSONData(data); 对此: responseGenerator.setJSONData(data.toMap()); 我希望这应该有效. P.S.:我的建议是删除JSONObject转换,而是返回一个实际类的Object,因为内部spring使用jackson,这是更强大的JSON框架然后org.json (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |