解決fastjson泛型轉換報錯的解決方法
阿新 • • 發佈:2020-11-24
Exception in thread "main" java.lang.ClassCastException: com.alibaba.fastjson.JSONObject cannot be cast to com.xh.demo.UserDO
泛性類
@Data public class ResultSetDTO<T> { private Integer totalSize; private Integer count; private List<T> records; }
實體類
@Data public class UserDO { private String id; private String name; }
Demo測試
public class AppDemo { public static void main(String[] args) { ResultSetDTO<UserDO> resultSetDTO = new ResultSetDTO<UserDO>(); resultSetDTO.setTotalSize(10); resultSetDTO.setCount(10); List<UserDO> list = new ArrayList<>(); UserDO userDO = null; for (int i = 10000; i < 10003; i++) { userDO = new UserDO(); userDO.setId(i + ""); userDO.setName("TEST_" + i); list.add(userDO); } resultSetDTO.setRecords(list); String jsonString = JSON.toJSONString(resultSetDTO); System.out.println(jsonString); ResultSetDTO<XsyUserDO> resultSet = JSON.parseObject(jsonString,ResultSetDTO.class); List<UserDO> records = resultSet.getRecords(); // 轉換是報異常 for (UserDO user : records) { System.out.println(user.getName()); } }
debuge 除錯時發現 resultSet.getRecords();
返回的是 ArrayList<JSONObject>
型別,不是 ArrayList<UserDO>
型別。由於型別引用必須強制指定目標物件,因此違背了 FastJson 規範,所以報:Exception in thread "main" java.lang.ClassCastException: com.alibaba.fastjson.JSONObject cannot be cast to com.xh.demo.UserDO。
解決方法
// 方法一 JSONObject parseObject = JSON.parseObject(jsonString); List<UserDO> records = JSON.parseArray(parseObject.getString("records"),UserDO.class); for (UserDO user : records) { System.out.println(user.getName()); } // 方法二 ResultSetDTO<XsyUserDO> resultSet = JSON.parseObject(jsonString,ResultSetDTO.class); List<UserDO> records = JSONObject.parseArray(resultSet.getRecords().toString(),UserDO.class); for (UserDO user : records) { System.out.println(user.getName()); } // 方法三 ResultSetDTO<XsyUserDO> resultSet = JSON.parseObject(jsonString,new TypeReference<ResultSetDTO<UserDO>>(){}); List<UserDO> records = resultSet.getRecords(); for (UserDO user : records){ System.out.println(user.getName());} }
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