关于Android定位获取详细地址的问题

时间:2022-10-07 13:42:02
遇到的问题是:运行在不同的手机上显示的地址不同,当然经纬度是一样的。请教各位大神,这是什么原因?

private Location getLocation(Context context){
        LocationManager manager = (LocationManager) context.getSystemService(context.LOCATION_SERVICE);
        //若GPS未开启
        if(!manager.isProviderEnabled(LocationManager.GPS_PROVIDER)){
            Toast.makeText(SigninActivity.this, "请开启GPS!", Toast.LENGTH_SHORT).show();
            Intent intent = new Intent(Settings.ACTION_LOCATION_SOURCE_SETTINGS);
            SigninActivity.this.startActivityForResult(intent, 0); //此为设置完成后返回到获取界面
        }
        //获得GPS支持
        location = manager.getLastKnownLocation(LocationManager.GPS_PROVIDER);
        Log.v("location1", location + "");
        if (location == null){
            //获得PASSIVE支持
            location = manager.getLastKnownLocation(LocationManager.PASSIVE_PROVIDER);
            Log.v("location2",location+"");
        }else {
            //获得NETWORK支持
            location = manager.getLastKnownLocation(LocationManager.NETWORK_PROVIDER);
            Log.v("location3",location+"");

        }

        geocoder = new Geocoder(SigninActivity.this);
        try {
            List<Address> list = geocoder.getFromLocation(location.getLatitude(),location.getLongitude(),5);
            Log.v("location",location.getLatitude()+","+location.getLongitude());
            if (list != null){
                for (int i=0; i<list.size(); i++){
                    Address address = list.get(i);
                    String add="";
                    int maxLine = address.getMaxAddressLineIndex();
                    Log.v("maxline",maxLine+"");
                    if (maxLine >= 2) {
                        add = address.getAddressLine(1) + address.getAddressLine(2);
                    } else {
                        add = address.getAddressLine(1);
                    }
                    tv_locaion.setText(add);
                }
            }
        } catch (IOException e) {
            e.printStackTrace();
        }
        return  location;
    }