C# 中的 Google 地理编码 Json 解析问题

Question

我的代码工作正常，但我似乎无法进入树的更深部分。我正在尝试拉经度和纬度。下面的代码将 'status' 拉为 'OK' 没问题（在响应的最后）。'geometry' -> 'location' -> 'lat' 和 'lng' 的语法是什么？

这是我的代码：

string RawAddress = "163 Leektown Road, New Gretna, NJ 08004";
string Address = RawAddress.Replace(" ", "+");
string AddressURL = "http://maps.google.com/maps/api/geocode/json?address=" + Address;
var result = new System.Net.WebClient().DownloadString(AddressURL);
dynamic data = JObject.Parse(result);

Lat.Text = data.status;

这是 API 生成的内容：

{
   "results" : [
  {
     "address_components" : [
        {
           "long_name" : "Mountain View",
           "short_name" : "Mountain View",
           "types" : [ "locality", "political" ]
        },
        {
           "long_name" : "Santa Clara County",
           "short_name" : "Santa Clara County",
           "types" : [ "administrative_area_level_2", "political" ]
        },
        {
           "long_name" : "California",
           "short_name" : "CA",
           "types" : [ "administrative_area_level_1", "political" ]
        },
        {
           "long_name" : "United States",
           "short_name" : "US",
           "types" : [ "country", "political" ]
        }
     ],
     "formatted_address" : "Mountain View, CA, USA",
     "geometry" : {
        "bounds" : {
           "northeast" : {
              "lat" : 37.4508789,
              "lng" : -122.0446721
           },
           "southwest" : {
              "lat" : 37.3567599,
              "lng" : -122.1178619
           }
        },
        "location" : {
           "lat" : 37.3860517,
           "lng" : -122.0838511
        },
        "location_type" : "APPROXIMATE",
        "viewport" : {
           "northeast" : {
              "lat" : 37.4508789,
              "lng" : -122.0446721
           },
           "southwest" : {
              "lat" : 37.3567599,
              "lng" : -122.1178619
           }
        }
     },
     "partial_match" : true,
     "types" : [ "locality", "political" ]
  }
  ],
  "status" : "OK"
  }

Answer 1

以下是获得所需内容的步骤：

1. 在http://json2csharp.com/ 中发布您的 JSON。获取生成的类并合并重复项，您会得到：

   public class AddressComponent
   {
       public string long_name { get; set; }
       public string short_name { get; set; }
       public List<string> types { get; set; }
   }
   
   public class Bounds
   {
       public Location northeast { get; set; }
       public Location southwest { get; set; }
   }
   
   public class Location
   {
       public double lat { get; set; }
       public double lng { get; set; }
   }
   
   public class Geometry
   {
       public Bounds bounds { get; set; }
       public Location location { get; set; }
       public string location_type { get; set; }
       public Bounds viewport { get; set; }
   }
   
   public class Result
   {
       public List<AddressComponent> address_components { get; set; }
       public string formatted_address { get; set; }
       public Geometry geometry { get; set; }
       public bool partial_match { get; set; }
       public List<string> types { get; set; }
   }
   
   public class RootObject
   {
       public List<Result> results { get; set; }
       public string status { get; set; }
   }
   

   （您也可以使用Paste JSON as Classes 或https://jsonutils.com/ 来生成初始类型定义。）

2. 使用Json.NET 反序列化您的 JSON，如下所示：

       var root = JsonConvert.DeserializeObject<RootObject>(result);
   

3. 您的查询返回了多个结果，因此您需要像这样遍历返回的位置：

       foreach (var singleResult in root.results)
       {
           var location = singleResult.geometry.location;
           var latitude = location.lat;
           var longitude = location.lng;
           // Do whatever you want with them.
       }
   

Answer 2

'geometry' -> 'location' -> 'lat' 和 'lng' 的语法是：

JObject data = JObject.Parse(result);

string lat = (string)data["results"][0]["geometry"]["location"]["lat"];
string lng = (string)data["results"][0]["geometry"]["location"]["lng"];