Result resDY = JSON.parseObject(msgStr,new TypeReference>() {});
我想把这句代码用工具类写出来,求怎么写?
Result类如下:
public class Result {
private String rspCode;
private String rspMsg;
private T result;
private List list;
}
Copyright 2014-2025 https://www.php.cn/ All Rights Reserved | php.cn | 湘ICP备2023035733号
相信楼主是需要这个一个方法声明,可以试试:
fastjson不是有实现吗