C＃中的Google Geocoding Json解析问题

小编典典

C＃中的Google Geocoding Json解析问题

json

我的代码工作正常，但似乎无法深入到树的更深部分。我正在尝试拉经度和纬度。下面的代码将“状态”毫无问题地拉为“确定”（在响应的最后），“几何”->“位置”->“
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;

API生成以下内容：

{
   "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"
  }

阅读 289

2020-07-27

共1个答案

小编典典

以下是获得所需内容的步骤：

将您的JSON发布到 http://json2csharp.com/中。采取结果类并合并重复项，您将得到：

        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; }
    }

（您还可以使用JSON作为类粘贴或 https://jsonutils.com/来生成初始类型定义。）

使用Json.NET反序列化JSON，如下所示：

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

您的查询返回了多个结果，因此您需要像这样循环遍历返回的位置：

    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.
}

2020-07-27