谷歌地图挂在慢速互联网连接,Android

时间:2015-06-07 08:37:30

标签: android android-fragments android-googleapiclient

在我的一项活动中,我正在片段中显示谷歌地图。它可以在3g或高速互联网连接下正常工作。但是在2g或非常慢的互联网连接中它会挂起我的手机,过了一段时间我就没有回复了#39;信息。 有什么方法可以处理这种情况吗?

这是我的工作代码: -

 public class MapLoader extends FragmentActivity implements View.OnClickListener {
  // GUI Widget
TextView lblNumber;
Button btnCall;
String number;
GoogleMap googleMap;
Bitmap myBitmap;
private String filePath;
/** Called when the activity is first created. */
@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.mapload);

    // Getting reference to the SupportMapFragment of activity_main.xml
    SupportMapFragment mapFragment = (SupportMapFragment) getSupportFragmentManager().findFragmentById(R.id.map);

    // Getting GoogleMap object from the fragment
    googleMap = mapFragment.getMap();
    googleMap.clear();
    String filterAddress = "";
    double lat=0;
    double lng=0;
    String ontime="";
    try {
        UserFunctions userFunction = new UserFunctions();
        JSONObject json = userFunction.getUserData(number);
        if(json.has("user")) {
            JSONArray android_version_array = json.getJSONArray("user");
            //getting android version
            for (int i = 0; i < android_version_array.length(); i++) {
                // creating new HashMap
                HashMap<String, String> map = new HashMap<String, String>();
                JSONObject myObj = android_version_array.getJSONObject(i);
                String lattiude = myObj.getString("latitude");
                String logitude = myObj.getString("longitude");
                ontime=myObj.getString("gpsTime");
                lat = Double.parseDouble(lattiude);
                lng = Double.parseDouble(logitude);
                // adding each child node to HashMap key => value
            }

            Geocoder geoCoder = new Geocoder(
                    getBaseContext(), Locale.getDefault());
            try {
                List<Address> addresses = geoCoder.getFromLocation(
                        lat,
                        lng, 1);

                if (addresses.size() > 0) {
                    for (int index = 0;
                         index < addresses.get(0).getMaxAddressLineIndex(); index++)
                        filterAddress += addresses.get(0).getAddressLine(index) + " ";

                }
                //  SupportMapFragment mapFragment = (SupportMapFragment) getSupportFragmentManager().findFragmentById(R.id.map);
                //  mapFragment.getView().findViewById()
                TextView a = (TextView) findViewById(R.id.test);
                a.setText(filterAddress);
            } catch (IOException ex) {
                ex.printStackTrace();
            } catch (Exception e2) {
                // TODO: handle exception

                e2.printStackTrace();
            }

             // Creating an instance of MarkerOptions to set position
            MarkerOptions markerOptions = new MarkerOptions();

            LatLng LOCATION = new LatLng(lat, lng);
            // Setting position on the MarkerOptions
            markerOptions.position(LOCATION);

            // Animating to the currently touched position
            // googleMap.animateCamera(CameraUpdateFactory.newLatLng(LOCATION),17.0f);
            googleMap.animateCamera(CameraUpdateFactory.newLatLngZoom(new LatLng(lat, lng), 17.0f));

            // Adding marker on the GoogleMap
            Marker marker = googleMap.addMarker(markerOptions);
            // Showing InfoWindow on the GoogleMap
            marker.showInfoWindow();

   }catch (Exception e){
        e.printStackTrace();
     }
  }
 }

注意: - 它不是完整的代码。出于安全原因,我删除了一些代码。

1 个答案:

答案 0 :(得分:2)

getMap()已弃用。您应该使用getMapAsync()并避免在UI线程上等待。