8

我正在编写一个 Android 应用程序,以将动态可用的纬度和对数坐标转换为人类可读的位置。

例如,12.2、4.5 位于英国伦敦市中心。关于粒度,我希望能够至少定位城市->城镇。或者如果不是,至少是城市,

有人可以就此问题提供哪些解决方案提供建议。

4

3 回答 3

10

试试这个:

//listenner location changed
private class MyLocListener implements LocationListener {
   public void onLocationChanged(Location location) {
      if (location != null) {
         Log.d("LOCATION CHANGED", location.getLatitude() + "");
         Log.d("LOCATION CHANGED", location.getLongitude() + "");
      }
   }
}

 //Get address base on location
try{
 Geocoder geo = new Geocoder(youractivityclassname.this.getApplicationContext(), Locale.getDefault());
 List<Address> addresses = geo.getFromLocation(latitude, longitude, 1);
  if (addresses.isEmpty()) {
        yourtextfieldname.setText("Waiting for Location");
  }
  else {
     if (addresses.size() > 0) {       
        Log.d(TAG,addresses.get(0).getFeatureName() + ", 
         " + addresses.get(0).getLocality() +", 
         " + addresses.get(0).getAdminArea() + ",
         " + addresses.get(0).getCountryName());

     }
  }
}
catch (Exception e) {
    e.printStackTrace(); 
}
于 2012-08-24T03:47:51.643 回答
2

将点位置(纬度、经度)转换为可读地址或地名的过程称为反向地理编码。[来自维基百科]

您必须使用GeoCoder类并使用方法getFromLocation。此方法返回List<Address>,您可以通过迭代列表中的每个Address对象来访问它。

例子:

  1. http://www.edumobile.org/android/android-development/gecoding-example/
  2. Android:反向地理编码 - getFromLocation
于 2012-08-24T03:59:41.040 回答
0

我认为这会产生更好的结果:

 private String convertLocationToAddress(Location location) {
    String addressText;
    String errorMessage = "";

    Geocoder geocoder = new Geocoder(getContext(), Locale.getDefault());

    List<Address> addresses = null;

    try {
        addresses = geocoder.getFromLocation(
                location.getLatitude(),
                location.getLongitude(),
                1
        );
    } catch (IOException ioException) {
        // Network or other I/O issues
        errorMessage = getString(R.string.network_service_error);
        Log.e(TAG, errorMessage, ioException);
    } catch (IllegalArgumentException illegalArgumentException) {
        // Invalid long / lat
        errorMessage = getString(R.string.invalid_long_lat);
        Log.e(TAG, errorMessage + ". " +
                "Latitude = " + location.getLatitude() +
                ", Longitude = " +
                location.getLongitude(), illegalArgumentException);
    }

    // No address was found
    if (addresses == null || addresses.size() == 0) {
        if (errorMessage.isEmpty()) {
            errorMessage = getString(R.string.no_address_found);
            Log.e(TAG, errorMessage);
        }
        addressText = errorMessage;

    } else {
        Address address = addresses.get(0);
        ArrayList<String> addressFragments = new ArrayList<>();

        // Fetch the address lines, join them, and return to thread
        for (int i = 0; i <= address.getMaxAddressLineIndex(); i++) {
            addressFragments.add(address.getAddressLine(i));
        }
        Log.i(TAG, getString(R.string.address_found));
        addressText =
                TextUtils.join(System.getProperty("line.separator"),
                        addressFragments);
    }

    return addressText;

}
于 2017-09-15T14:17:42.820 回答