Question

我想将我的JsonObject转换为纬度和经度的ArrayList。三天我一直在努力解决这个问题而没有任何结果。请帮忙！

这是我的JSON：

{
   "hits":[
      {
         "_geoloc":{
            "lat":33.842624,
            "lng":-5.480229
         },
         "objectID":"-KsPQUlvJK-PCSrH3Dq3"
      },
      {
         "_geoloc":{
            "lat":33.84878,
            "lng":-5.481698
         },
         "objectID":"-KsPQUloi4BAduevPJta"
      }
   ],
   "nbHits":2,
   "page":0,
   "nbPages":1,
   "hitsPerPage":20,
   "processingTimeMS":1,
   "exhaustiveNbHits":true,
   "query":"",
   "params":"aroundLatLng=33.62727138006218%2C-4.498191252350807&aroundRadius=545000"
}

这是我的代码：

client.getIndex("users").searchAsync(query, new CompletionHandler() {
    @Override
    public void requestCompleted(JSONObject jsonObject, AlgoliaException e) {
    }
});

Answer 1

您可以使用JsonObject的成员方法执行此操作（http://docs.oracle.com/javaee/7/api/javax/json/JsonObject.html）。例如，类似以下内容：

class GeoPoint {
    private double lat;
    private double lon;
    public GeoPoint( double lat, double lon ) {
        this.lat = lat
        this.lon = lon
    }
    // ...
}
ArrayList<GeoPoint> list = new ArrayList<>();
// Thankfully JsonArray objects are iterable...
for (JsonValue e : jsonObject.getJsonArray("hits")) {
    JsonObject coordWrapper = (JsonObject) e;
    JsonObject coord = coordWrapper.getJsonObject("_geoloc");
    double lat = coord.getJsonNumber("lat").doubleValue();
    double lon = coord.getJsonNumber("lon").doubleValue();
    list.add(new GeoPoint(lat, lon));
}

Answer 2

如果您使用fastjson作为JSON库，可以这样做：

public class GeoLoc {
    private double lng;
    private double lat;
    // getter & setter...
}

public class Geo {
    private String objectID;

    @JSONField(name = "_geoloc")
    private GeoLoc geoLoc;
    // getter & setter...
}

public static void main(String[] args) {
    String jsonStr = ""; // Your json string here.
    JSONObject jsonObject = JSON.parseObject(json);
    JSONArray jsonArray = jsonObject.getJSONArray("hits");
    List<Geo> geos = jsonArray.toJavaList(Geo.class);
    List<GeoLoc> geoLocs = geos.stream()
            .map(Geo::getGeoLoc)
            .collect(Collectors.toList());
}

不需要显式转换为List，fastjson会为你做这件事。

将JSONObjects转换为ArrayList

2 个答案: