SimpleJSON how the get "location" item values in JSON array - arrays

Question

"results" : [
{
"geometry" : {
"location" : {
"lat" : 50.4501,
"lng" : 30.5234
},
"viewport" : {
"northeast" : {
"lat" : 50.59079800991073,
"lng" : 30.82594104187906
},
"southwest" : {
"lat" : 50.21327301525928,
"lng" : 30.23944009690609
}
}
},
There are lat and lng values in geometry, but i can get only into "results" for now.
JSONNode root = JSONNode.Parse(www.downloadHandler.text);
JSONArray nodes = root["results"].AsArray;

Answer 1

Might be better to use JsonElement which is faster. I have tested this code against your json and it works.
public static double UsingJsonElement(string jsonString)
{
var jsonElement = JsonSerializer.Deserialize<JsonElement>(jsonString);
return jsonElement
.GetProperty("geometry")
.GetProperty("location").GetProperty("lat")
.GetDouble();
var viewportlat = jsonElement
.GetProperty("geometry")
.GetProperty("viewport").GetProperty("northeas").GetProperty("lat")
.GetDouble();
}

Answer 2

According to your JSON hierarchy I guess it would depend on exactly which set of lat/long you are interested in - there are three.
And then afaik you simply go down the hierarchy like e.g.
var firstNode = nodes[0];
var geometry = firstNode["geometry"];
var location = geometry["location"];
var lat = location["lat"].AsFloat;
var lng = location["lng"].AsFloat;
You might want to rather use Newtonsoft JSON.Net though and simply implement according target type
[Serializable]
public class Root
{
public Node[] results;
}
[Serializable]
public class Node
{
// again depending which values you are interested in
// you only have to implement th hierarchy of values you care for
// the rest will be ignored
public Geometry geometry;
}
[Serializable]
public class Geometry
{
public Location location;
}
[Serializable]
public class Location
{
public float lat;
public float lng;
}
and then simply go
var nodes = JsonConvert.DeserializeObject<Root>(jsonString).results;
and access
var lat = nodes[0].geometry.location.lat;
var lng = nodes[0].geometry.location.lng;