24

如何在Android中获取国家名称，如果已知地理坐标？如何使其成为最简单的方法？

4

4 回答 4

30

干得好

Geocoder gcd = new Geocoder(context, Locale.getDefault());
List<Address> addresses = gcd.getFromLocation(lat, lng, 1);

    if (addresses.size() > 0)
     {  
       String countryName=addresses.get(0).getCountryName();
     }
于 2012-06-18T12:20:32.053 回答
23

为此使用下面的功能。

public void getAddress(double lat, double lng) {
    Geocoder geocoder = new Geocoder(HomeActivity.mContext, Locale.getDefault());
    try {
        List<Address> addresses = geocoder.getFromLocation(lat, lng, 1);
        Address obj = addresses.get(0);
        String add = obj.getAddressLine(0);
        GUIStatics.currentAddress = obj.getSubAdminArea() + ","
                + obj.getAdminArea();
        GUIStatics.latitude = obj.getLatitude();
        GUIStatics.longitude = obj.getLongitude();
        GUIStatics.currentCity= obj.getSubAdminArea();
        GUIStatics.currentState= obj.getAdminArea();
        add = add + "\n" + obj.getCountryName();
        add = add + "\n" + obj.getCountryCode();
        add = add + "\n" + obj.getAdminArea();
        add = add + "\n" + obj.getPostalCode();
        add = add + "\n" + obj.getSubAdminArea();
        add = add + "\n" + obj.getLocality();
        add = add + "\n" + obj.getSubThoroughfare();

        Log.v("IGA", "Address" + add);
        // Toast.makeText(this, "Address=>" + add,
        // Toast.LENGTH_SHORT).show();

        // TennisAppActivity.showDialog(add);
    } catch (IOException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
        Toast.makeText(this, e.getMessage(), Toast.LENGTH_SHORT).show();
    }
}

在清单中添加以下权限

<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/>
于 2012-06-18T12:25:23.973 回答
8

用这个

try {
        Geocoder geo = new Geocoder(this.getApplicationContext(), Locale.getDefault());
        List<Address> addresses = geo.getFromLocation(location.getLatitude(), location.getLongitude(), 1);
        if (addresses.isEmpty()) {
            placeName.setText("Waiting for Location");
        }
        else {
            if (addresses.size() > 0) {
                placeName.setText(addresses.get(0).getFeatureName() + ", " + addresses.get(0).getLocality() +", " + addresses.get(0).getAdminArea() + ", " + addresses.get(0).getCountryName());
            }
        }
    }
    catch(Exception e){
        Toast.makeText(this, "No Location Name Found", 600).show();
    }

在清单文件中使用它

<uses-permission android:name="android.permission.ACCESS_FINE_LOCATION"/>
于 2012-06-18T12:28:06.403 回答
2

我最近的做法是从 Web API URL 读取数据并解析 JSON。

点 (40.714224, -73.961452) 的示例 URL 是：

http://maps.google.com/maps/geo?q=40.714224,-73.961452&output=json&oe=utf8&sensor=true_or_false&key=your_api_key

产生以下输出：

{
  "name": "40.714224,-73.961452",
  "Status": {
    "code": 200,
    "request": "geocode"
  },
  "Placemark": [ {
    "id": "p1",
    "address": "285 Bedford Ave, Brooklyn, NY 11211, USA",
    "AddressDetails": {
   "Accuracy" : 8,
   "Country" : {
      "AdministrativeArea" : {
         "AdministrativeAreaName" : "NY",
         "SubAdministrativeArea" : {
            "Locality" : {
               "DependentLocality" : {
                  "DependentLocalityName" : "Williamsburg",
                  "PostalCode" : {
                     "PostalCodeNumber" : "11211"
                  },
                  "Thoroughfare" : {
                     "ThoroughfareName" : "285 Bedford Ave"
                  }
               },
               "LocalityName" : "Brooklyn"
            },
            "SubAdministrativeAreaName" : "Kings"
         }
      },
      "CountryName" : "USA",
      "CountryNameCode" : "US"
   }
},
    "ExtendedData": {
      "LatLonBox": {
        "north": 40.7154779,
        "south": 40.7127799,
        "east": -73.9600584,
        "west": -73.9627564
      }
    },
    "Point": {
      "coordinates": [ -73.9614074, 40.7141289, 0 ]
    }
  } ]
}

我发现 GSON 非常适合在 Android 中解析 JSON。

于 2012-06-18T12:29:08.430 回答