object convert to json how to modify the property key? - json

Question

i have a json like
"weatherInfo":{
"city":"北京",
"publishDate":"2014年3月4日",
"week":"星期二",
"tempRange":"8℃~-3℃",
"feelTemp":"10",
"time":"16:05",
"temp":"11",
"WD":"北风",
"WS":"2级",
"SD":"27%",
"weather1":"晴"
}
and my class is
public class WeatherVO implements Serializable{
private static final long serialVersionUID = 2348480036959754071L;
#JsonProperty(value="weatherinfo")
private WeatherInfoVO weatherInfoVO;
#JsonIgnoreProperties(ignoreUnknown=true)
public class WeatherInfoVO{
//城市
#JsonProperty(value="city")
private String city;
//发布日期
private String publishDate;
//发布时间
#JsonProperty(value="time")
private String publishTime;
//星期
private String week;
//温度范围
private String tempRange;
//当前时刻温度
#JsonProperty(value="temp")
private String currentTemp;
//风向
#JsonProperty(value="WD")
private String windDirection;
//风力
#JsonProperty(value="WS")
private String windForce;
//当前时刻湿度
#JsonProperty(value="SD")
private String currentHumidity;
//体感温度
private String feelTemp;
//天气描述
private String weatherDesc;
}
}
i want to convert json to object like:
WeatherVO weatherVO = objectMapper.readValue (jsonString, WeatherVO.class);
and i want to return json use org.springframework.http.converter.json.MappingJacksonHttpMessageConverter class
and the return json is
"weatherInfo":{
"city":"北京",
"publishDate":"2014年3月4日",
"week":"星期二",
"tempRange":"8℃~-3℃",
"feelTemp":"10",
"time":"16:05",
"temp":"11",
"WD":"北风",
"WS":"2级",
"SD":"27%",
"weather1":"晴"
}
but i want to return like
"weatherInfo":{
"city":"北京",
"publishDate":"2014年3月4日",
"week":"星期二",
"tempRange":"8℃~-3℃",
"feelTemp":"10",
"publishTime":"16:05",
"currentTemp":"11",
"windDirection":"北风",
"windForce":"2级",
"currentHumidity":"27%",
"weather":"晴"
}
how can i do that?
thanks for your any suggestion and help.

Answer 1

You can achieve this by marking the getter of the field by #JsonIgnore then create another method that get the field and mark it as #JsonProperty.
My advice is to rename first your fields to avoid confusion. I renamed it like the json you have. So for example.
public class WeatherInfoVO{
private String city;
private String publishDate;
private String week;
private String tempRange;
private String feelTemp;
private String time;
private String temp;
private String WD;
private String WS;
private String SD;
private String weather1;
//getters and setters
}
then in you setters method mark the appropriate #JsonProperty.
#JsonProperty(value="time")
public void setTime(String time) {
this.temp = temp;
}
#JsonProperty(value="temp")
public void setTemp(String temp) {
this.temp = temp;
}
#JsonProperty(value="WD")
public void setWD(String WD) {
this.WD = WD;
}
#JsonProperty(value="WS")
public void setWS(String WS) {
this.WS = WS;
}
#JsonProperty(value="SD")
public void setSD(String SD) {
this.SD = SD;
}
#JsonProperty(value="weather1")
public void setWeather1(String weather1) {
this.weather1 = weather1;
}
//other setters here
and in your getters, make sure to add #JsonIgnore to the fields you want to rename. Since you declare it as #JsonIgnore, you need to create another getter and mark it as #JsonProperty. Do it only to the fields you are going to rename, in your case the fields are time, temp, WD, WS, SD and weather1 only. Here's an example.
#JsonIgnore
public void getTime(){
return time;
}
#JsonProperty(value="publishTime")
public void getPublishTime(){
return time;
}
#JsonIgnore
public void getTemp(){
return temp;
}
#JsonProperty(value="currentTemp")
public void getCurrentTemp(){
return temp;
}
#JsonIgnore
public void getWD(){
return WD;
}
#JsonProperty(value="windDirection")
public void getWindDirection(){
return WD;
}
#JsonIgnore
public void getWS(){
return WS;
}
#JsonProperty(value="windForce")
public void getWindForce(){
return WS;
}
#JsonIgnore
public void getSD(){
return SD;
}
#JsonProperty(value="currentHumidity")
public void getCurrentHumidity(){
return SD;
}
#JsonIgnore
public void getWeather1(){
return weather1;
}
#JsonProperty(value="weather")
public void getWeather(){
return weather1;
}