将纬度和经度转换为esri arcGIS MapPoint

Question

我在将纬度和经度值转换为android esri arcGIS map Point时遇到了麻烦.这是我从GPS坐标获取纬度和经度值的代码:

LocationManager lm;
String towers;
double lat;
double longi;
TextView txt;

            lm = (LocationManager) getSystemService(Context.LOCATION_SERVICE);
            Criteria crit = new Criteria();
            towers = lm.getBestProvider(crit, false);
            Location location = lm.getLastKnownLocation(towers);

            if(location != null)
            {
                lat = location.getLatitude();
                longi = location.getLongitude();
            }

现在我有纬度和经度值.现在我需要的是将这些值转换为有效的esri arcGIS MapPoint.谁能帮我？

提前致谢.

Answer 1

假设您正在使用ESRI Android API？如果是这样,请在地图上创建图形图层.然后创建一个点对象

com.esri.core.geometry.Point
Point myPoint = new Point();

然后设置x/y值:

myPoint.setX(longi);
myPoint.setY(lat);

然后将myPoint添加到图形对象.

http://help.arcgis.com/en/arcgismobile/10.0/apis/android/api/index.html

Answer 2

对的,这是可能的.但是你没有在ArcGis中使用locationmanager.

ArcGIS具有预定义的方法,如LocationListener,即:OnStatusChangedListener.

请参阅以下代码,将位置纬度和经度转换为esri arcGIS MapPoint.

     mMapView.setOnStatusChangedListener(new OnStatusChangedListener() {

            /**
             * 
             */
      private static final long serialVersionUID = 1L;

      public void onStatusChanged(Object source, STATUS status) {
      if (source == mMapView && status == STATUS.INITIALIZED) {
      LocationService ls = mMapView.getLocationService();
      ls.setAutoPan(false);
      ls.setLocationListener(new LocationListener() {

      boolean locationChanged = false;

      // Zooms to the current location when first GPS fix
      // arrives.
      public void onLocationChanged(Location loc) {
      if (!locationChanged) {
      locationChanged = true;
      double locy = loc.getLatitude();
      double locx = loc.getLongitude();
      Point wgspoint = new Point(locx, locy);
      Point mapPoint = (Point) GeometryEngine.project(wgspoint,  

      SpatialReference.create(4326),

      mMapView.getSpatialReference());

      Unit mapUnit = mMapView.getSpatialReference().getUnit();
      double zoomWidth = Unit.convertUnits(

      SEARCH_RADIUS, Unit.create(LinearUnit.Code.MILE_US), mapUnit);
      Envelope zoomExtent = new Envelope(mapPoint, zoomWidth, zoomWidth);

      mMapView.setExtent(zoomExtent);

      GraphicsLayer gLayer = new GraphicsLayer();
      PictureMarkerSymbol symbol = new     
      PictureMarkerSymbol(getResources().getDrawable(R.drawable.twiz_car_red));
      Graphic graphic = new Graphic(mapPoint, symbol);
      //Graphic point=new Graphic(new Point(x, y),new    
      SimpleMarkerSymbol(Color.CYAN,20,STYLE.CIRCLE));   
      gLayer.addGraphic(graphic);
      mMapView .addLayer(gLayer);

         }
      }

      public void onProviderDisabled(String arg0) {

            }
      public void onProviderEnabled(String arg0) {
            }

      public void onStatusChanged(String arg0, int arg1,
      Bundle arg2) {

         }
        });
      ls.start();

     }
   }
});

Answer 3

我从这里借了一些代码

private Point ToGeographic(Point pnt)
{
    double mercatorX_lon = pnt.getX();
    double mercatorY_lat = pnt.getY();
    if (Math.abs(mercatorX_lon) < 180 && Math.abs(mercatorY_lat) < 90)
        return pnt;

    if ((Math.abs(mercatorX_lon) > 20037508.3427892) || (Math.abs(mercatorY_lat) > 20037508.3427892))
        return pnt;

    double x = mercatorX_lon;
    double y = mercatorY_lat;
    double num3 = x / 6378137.0;
    double num4 = num3 * 57.295779513082323;
    double num5 = Math.floor((double)((num4 + 180.0) / 360.0));
    double num6 = num4 - (num5 * 360.0);
    double num7 = 1.5707963267948966 - (2.0 * Math.atan(Math.exp((-1.0 * y) / 6378137.0)));
    mercatorX_lon = num6;
    mercatorY_lat = num7 * 57.295779513082323;
    return new Point(mercatorX_lon, mercatorY_lat);
}

private Point ToWebMercator(Point pnt)
{
    double mercatorX_lon = pnt.getX();
    double mercatorY_lat = pnt.getY();
    if ((Math.abs(mercatorX_lon) > 180 || Math.abs(mercatorY_lat) > 90))
        return pnt;

    double num = mercatorX_lon * 0.017453292519943295;
    double x = 6378137.0 * num;
    double a = mercatorY_lat * 0.017453292519943295;

    mercatorX_lon = x;
    mercatorY_lat = 3189068.5 * Math.log((1.0 + Math.sin(a)) / (1.0 - Math.sin(a)));
    return new Point(mercatorX_lon, mercatorY_lat);
}

我没有提出效率,但它至少是一个起点.

Answer 4

免责声明：我不是这方面的专家，但想尽力提供帮助。:)

现在有一个ArcGIS Stack Exchange站点。一直在添加更多信息，与互联网上发布的信息相比，这是一个很好的综合资源。

对于框架，我推荐GeoTools for Android。

顺便说一句，QGIS for Android是 Marco Bernasocchi 的一个有趣项目，您可能会发现它作为参考很有帮助。

希望您能找到您要找的东西！