谷歌地理编码Json解析C#中的问题

时间:2015-02-06 17:25:57

标签: c# json api geocoding

我的代码运行正常,但我似乎无法进入树的更深层部分。我试图拉经度和纬度。下面的代码将“状态”没有问题提取为“OK”(在响应的最后)。“几何”的语法是什么 - >'location' - >'lat'和'lng'?

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;

这就是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"
  }

2 个答案:

答案 0 :(得分:16)

以下是获得所需内容的步骤:

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

    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 Classeshttps://jsonutils.com/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,因此您需要循环浏览返回的位置,如下所示:

        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.
        }
    

答案 1 :(得分:0)

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

JObject data = JObject.Parse(result);

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