解决fastjson泛型转换报错的解决方法

错误信息

Exception in thread "main" java.lang.ClassCastException: com.alibaba.fastjson.JSONObject cannot be cast to com.xh.demo.UserDO

泛性类

@Data
public class ResultSetDTO<T> {
 private Integer totalSize;
 private Integer count;
 private List<T> records;
}

实体类

@Data
public class UserDO {

private String id;
 private String name;
}

Demo测试

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();
   // 转换是报异常
   for (UserDO user : records) {
     System.out.println(user.getName());
   }
 }

debuge 调试时发现 resultSet.getRecords(); 返回的是 ArrayList<JSONObject> 类型，不是 ArrayList<UserDO> 类型。由于类型引用必须强制指定目标对象，因此违背了 FastJson 规范，所以报：Exception in thread "main" java.lang.ClassCastException: com.alibaba.fastjson.JSONObject cannot be cast to com.xh.demo.UserDO。

解决方法

// 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());}
}

来源：https://blog.csdn.net/xhaimail/article/details/109890503