无法生成GPS位置

时间:2013-12-27 04:09:22

标签: android

protected JSONObject doWork() {

        Location location;
        Double latitude = 0.0;
 locationManager = (LocationManager) getSystemService(Context.LOCATION_SERVICE);

        isGPSEnabled = locationManager
                .isProviderEnabled(LocationManager.GPS_PROVIDER);

        // getting network status
        isNetworkEnabled = locationManager
                .isProviderEnabled(LocationManager.NETWORK_PROVIDER);
        if (!isGPSEnabled && !isNetworkEnabled) {
            // no network provider is enabled
        }

        else {
            this.canGetLocation = true;
            // First get location from Network Provider
            if (isNetworkEnabled) {
                locationManager.requestLocationUpdates(
                        LocationManager.NETWORK_PROVIDER, MIN_TIME_BW_UPDATES,
                        MIN_DISTANCE_CHANGE_FOR_UPDATES, this);
                Log.d("Network", "Network");
                if (locationManager != null) {
                    location = locationManager
                            .getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
                    if (location != null) {
                        Log.i("Provider", provider + " has been selected.");
                        latitude = location.getLatitude();
                        onLocationChanged(location);
                    }

                }
                // latitude = location.getLatitude();
                // longitude = location.getLongitude();
            }
        }
        JSONObject result = new JSONObject();
           try {

            String msg = "KAsun " + mHelloTo + " - its currently "
                    + latitude.toString();
            result.put("Message", msg);

            Log.d(TAG, msg);

        } catch (JSONException e) {
        }

        return result;

我尝试上面的代码通过NETWORK生成当前的GPS位置。但是当我尝试我得到以下错误。我习惯这个代码作为后台服务。你能告诉我为什么会出现这个错误

 12-27 09:34:39.406: I/BackgroundService(14495): java.lang.RuntimeException: Can't create handler inside thread that has not called Looper.prepare()

0 个答案:

没有答案