我已经学会了如何通过Retrofit Gson解析JSON对象,但我需要通过Retrofit Gson解析一个完整的JSON数组。
我需要解析以下内容:
"{\n" +
" \"snappedPoints\": [\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.2784167,\n" +
" \"longitude\": 149.1294692\n" +
" },\n" +
" \"originalIndex\": 0,\n" +
" \"placeId\": \"ChIJoR7CemhNFmsRQB9QbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.280321693840129,\n" +
" \"longitude\": 149.12908274880189\n" +
" },\n" +
" \"originalIndex\": 1,\n" +
" \"placeId\": \"ChIJiy6YT2hNFmsRkHZAbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.280960897210818,\n" +
" \"longitude\": 149.1293250692261\n" +
" },\n" +
" \"originalIndex\": 2,\n" +
" \"placeId\": \"ChIJW9R7smlNFmsRMH1AbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.28142839817933,\n" +
" \"longitude\": 149.1298619971291\n" +
" },\n" +
" \"originalIndex\": 3,\n" +
" \"placeId\": \"ChIJy8c0r2lNFmsRQEZUbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.28193988170618,\n" +
" \"longitude\": 149.13001013387623\n" +
" },\n" +
" \"originalIndex\": 4,\n" +
" \"placeId\": \"ChIJ58xCoGlNFmsRUEZUbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.282819705480151,\n" +
" \"longitude\": 149.1295597114644\n" +
" },\n" +
" \"originalIndex\": 5,\n" +
" \"placeId\": \"ChIJabjuhGlNFmsREIxAbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.283139388422363,\n" +
" \"longitude\": 149.12895618087012\n" +
" },\n" +
" \"originalIndex\": 6,\n" +
" \"placeId\": \"ChIJ1Wi6I2pNFmsRQL9GbW7qABM\"\n" +
" },\n" +
" {\n" +
" \"location\": {\n" +
" \"latitude\": -35.284728724835304,\n" +
" \"longitude\": 149.12835061713685\n" +
" },\n" +
" \"originalIndex\": 7,\n" +
" \"placeId\": \"ChIJW5JAZmpNFmsRegG0-Jc80sM\"\n" +
" }\n" +
" ]\n" +
"}"
我只需要lat和long。
这是我通过Retrofit GSON解析JSON对象的方法
package com.example.akshay.retrofitgson;
import com.google.gson.annotations.Expose;
import com.google.gson.annotations.SerializedName;
/**
* Created by Akshay on 9/6/2015.
*/
public class gitmodel {
@SerializedName("latitude")
@Expose
public String latitude;
@SerializedName("longitude")
@Expose
public String longitude;
public void setLatitude(String latitude)
{
this.latitude = latitude;
}
public String getLatitude()
{
return latitude;
}
public void setLongitude(String longitude)
{
this.longitude = longitude;
}
public String getlatitude()
{
return latitude;
}
}
答案 0 :(得分:0)
Retrofit使用Gson library来序列化Json响应。只要您正确设置了模型类,就可以告诉Gson尝试将json序列化为该模型类的实例。您的代码应如下所示:
String json; // retrieved from file/server
MyObject myObject = new Gson().fromJson(json, MyObject.class)
答案 1 :(得分:0)
您可能只需要lat / longs,但最简单的方法是设置POJO以获取所有内容,然后从POJO中提取lat long。如果需要,您甚至可以设计反序列化对象以隐藏内部对象。对于你的JSON,这很容易,只需这样做:
public static class SnappedPoints {
private List<Point> snappedPoints;
public String toString() {
return snappedPoints.toString();
}
}
public static class Point {
private Location location;
public double getLatitude() {
return location.getLatitude();
}
public double getLongitude() {
return location.getLongitude();
}
public String toString() {
return "{" + location.getLatitude() + "," + location.getLongitude() + "}";
}
}
public static class Location {
double latitude;
double longitude;
public double getLatitude() {
return latitude;
}
public double getLongitude() {
return longitude;
}
}
只需执行此操作即可看到此操作:
public static void main(String[] args) {
System.out.println(new Gson().fromJson(json, SnappedPoints.class));
}
或者,在Retrofit的情况下,这样的事情:
public interface MyRetrofitAPI {
@GET("/path/to/json")
void getSnappedPoints(/* arguments */, Callback<SnappedPoints> callback);
}