我想从纬度和经度在安卓得到以下值
街道地址 城市/州 邮政编码 完整的地址
如何做到这一点?
我想从纬度和经度在安卓得到以下值
街道地址 城市/州 邮政编码 完整的地址
如何做到这一点?
当前回答
使用Geocoder类从纬度和经度中获得完整地址非常容易。下面是代码示例。希望这能有所帮助!
if (l != null) {
val lat = l.latitude
val lon = l.longitude
val geocoder = Geocoder(this, Locale.getDefault())
val addresses: List<Address>
addresses = geocoder.getFromLocation(lat, lon, 1)
val address = addresses[0].getAddressLine(0)
val address2 = addresses[0].getAddressLine(1)
val city = addresses[0].locality
val state = addresses[0].adminArea
val country = addresses[0].countryName
val postalCode = addresses[0].postalCode
val knownName = addresses[0].featureName
val message =
"Emergency situation. Call for help. My location is: " + address + "." + "http://maps.google.com/maps?saddr=" + lat + "," + lon
}
您可以只使用地址值,因为它提供了所有完整的地址。如果需要单独的组件,也可以使用其他组件。
其他回答
public String getAddress(LatLng latLng) {
String cAddress = "";
if (latLng == null) {
errorMessage = "no_location_data_provided";
Log.wtf(TAG, errorMessage);
return "";
}
Geocoder geocoder = new Geocoder(this, Locale.getDefault());
// Address found using the Geocoder.
List<Address> addresses = null;
try {
// Using getFromLocation() returns an array of Addresses for the area immediately
// surrounding the given latitude and longitude. The results are a best guess and are
// not guaranteed to be accurate.
addresses = geocoder.getFromLocation(
latLng.latitude,
latLng.longitude,
// In this sample, we get just a single address.
1);
} catch (IOException ioException) {
// Catch network or other I/O problems.
errorMessage = "service_not_available";
Log.e(TAG, errorMessage, ioException);
} catch (IllegalArgumentException illegalArgumentException) {
// Catch invalid latitude or longitude values.
errorMessage = "invalid_lat_long_used";
Log.e(TAG, errorMessage + ". " +
"Latitude = " + latLng.latitude +
", Longitude = " + latLng.longitude, illegalArgumentException);
}
// Handle case where no address was found.
if (addresses == null || addresses.size() == 0) {
if (errorMessage.isEmpty()) {
errorMessage = "no_address_found";
Log.e(TAG, errorMessage);
}
} else {
Address address = addresses.get(0);
ArrayList<String> addressFragments = new ArrayList<String>();
// Fetch the address lines using {@code getAddressLine},
// join them, and send them to the thread. The {@link android.location.address}
// class provides other options for fetching address details that you may prefer
// to use. Here are some examples:
// getLocality() ("Mountain View", for example)
// getAdminArea() ("CA", for example)
// getPostalCode() ("94043", for example)
// getCountryCode() ("US", for example)
// getCountryName() ("United States", for example)
String allAddress = "";
for (int i = 0; i < address.getMaxAddressLineIndex(); i++) {
addressFragments.add(address.getAddressLine(i));
allAddress += address.getAddressLine(i) + " ";
}
if (address.getAdminArea() != null) {
state = address.getAdminArea();
} else {
state = "";
}
if (address.getLocality() != null) {
city = address.getLocality();
} else {
city = "";
}
if (address.getPostalCode() != null) {
postalCode = address.getPostalCode();
} else {
postalCode = "";
}
Log.i(TAG, "address_found");
//driverAddress = TextUtils.join(System.getProperty("line.separator"), addressFragments);
cAddress = allAddress;
Log.e("result", cAddress.toString());
}
return cAddress;
}
您可以使用此方法对正确完整的地址进行地理编码
从latlong (Geo-coordinates)获取Address还有最后一个技巧。你可以简单地通过经度和纬度点击谷歌地图web服务。它只是一个GET-Method web服务。
它将返回JSON响应,可以很容易地解析以获得地址。它的URL是:
http://maps.googleapis.com/maps/api/geocode/json?latlng=32,75&sensor=true
你可以用lat,long代替32,75。
只要用这个方法,把你的背,长。
public static void getAddress(Context context, double LATITUDE, double LONGITUDE{
//Set Address
try {
Geocoder geocoder = new Geocoder(context, Locale.getDefault());
List<Address> addresses = geocoder.getFromLocation(LATITUDE, LONGITUDE, 1);
if (addresses != null && addresses.size() > 0) {
String address = addresses.get(0).getAddressLine(0); // If any additional address line present than only, check with max available address lines by getMaxAddressLineIndex()
String city = addresses.get(0).getLocality();
String state = addresses.get(0).getAdminArea();
String country = addresses.get(0).getCountryName();
String postalCode = addresses.get(0).getPostalCode();
String knownName = addresses.get(0).getFeatureName(); // Only if available else return NULL
Log.d(TAG, "getAddress: address" + address);
Log.d(TAG, "getAddress: city" + city);
Log.d(TAG, "getAddress: state" + state);
Log.d(TAG, "getAddress: postalCode" + postalCode);
Log.d(TAG, "getAddress: knownName" + knownName);
}
} catch (IOException e) {
e.printStackTrace();
}
return;
}
用这个对我有用:D
检索纬度和经度的json数据。
https://maps.googleapis.com/maps/api/geocode/json?key=AIzaSyAr29XeWWAeWZcrOgjjfs3iSnqkWtAz4No&latlng=2.1812,102.4266&sensor=true
用你自己的地方改变纬度、经度。
https://maps.googleapis.com/maps/api/geocode/json?key= < \ API_KEY_HERE > &latlng =“纬度,经度”&sensor = true
您可以使用自己的密钥更改<\API_KEY_HERE>。
需要在谷歌控制台为新的api密钥启用api服务。
D选项是正确答案
如果您使用Kotlin语言,我创建这个方法来直接获取地址位置
private fun getAddress(latLng: LatLng): String {
val geocoder = Geocoder(this, Locale.getDefault())
val addresses: List<Address>?
val address: Address?
var addressText = ""
addresses = geocoder.getFromLocation(latLng.latitude, latLng.longitude, 1)
if (addresses.isNotEmpty()) {
address = addresses[0]
addressText = address.getAddressLine(0)
} else{
addressText = "its not appear"
}
return addressText
}
但是当你调用这个方法时,这个方法只返回String值
如果你想获取所有地址,你只需使用这个方法/函数
fun getAddress(latLng: LatLng){
val geocoder = Geocoder(this, Locale.getDefault())
val addresses: List<Address>?
val address: Address?
var fulladdress = ""
addresses = geocoder.getFromLocation(latLng.latitude, latLng.longitude, 1)
if (addresses.isNotEmpty()) {
address = addresses[0]
fulladdress = address.getAddressLine(0) // If any additional address line present than only, check with max available address lines by getMaxAddressLineIndex
var city = address.getLocality();
var state = address.getAdminArea();
var country = address.getCountryName();
var postalCode = address.getPostalCode();
var knownName = address.getFeatureName(); // Only if available else return NULL
} else{
fulladdress = "Location not found"
}
}