【发布时间】:2021-04-21 20:19:42
【问题描述】:
如何正确实现dto类,让拿到json的时候不用解析?
比如对于这样一个json对象:
{"errorCode":"0","errorMessage":"Success","actionCode":71015,
"actionCodeDescription":"Operation declined. ",
"amount":100000,"date":1618750705018,
"OrderParams":[{"name":"Finish","value":"false"}],
"attributes":[{"name":"number","value":"6a883ef0"}],
"cardInfo":{"pan":"111111**1111","expiration":"202202"},
"id":"123456","auth":"110101010"}
您需要从中获取键的值: actionCode , 金额, 日期 , pan
更新
import com.fasterxml.jackson.annotation.JsonFormat;
import com.fasterxml.jackson.annotation.JsonProperty;
import lombok.Data;
@Data
public class OpenJsonFormat {
@JsonProperty("actionCode")
private String actionCode;
@JsonProperty("amount")
private String amount;
@JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "dd-MM-yyyy hh:mm:ss")
private Date date;
@JsonProperty("pan")
private String pan;
}
【问题讨论】:
-
对于那些名称与JSON字符串中对应字段名称相同的变量,您不需要添加
@JsonProperty。