函数返回 Future<dynamic>

Question

所以我正在学习颤振，我有一个返回 UserLocation 对象的函数 -

getUserLocation() async {
    bool _serviceEnabled;
    loc.PermissionStatus _permissionGranted;

    _serviceEnabled = await location.serviceEnabled();
    if (!_serviceEnabled) {
      _serviceEnabled = await location.requestService();
      if (!_serviceEnabled) {
        return;
      }
    }
    _permissionGranted = await location.hasPermission();
    if (_permissionGranted == loc.PermissionStatus.denied) {
      _permissionGranted = await location.requestPermission();
      if (_permissionGranted != loc.PermissionStatus.granted) {
        return;
      }
    }
    try {
      _currentPosition = await location.getLocation();
    } catch (e) {
      print(e);
    }
    List<geo.Placemark> placemarks = await geo.placemarkFromCoordinates(
        _currentPosition.latitude ?? 0, _currentPosition.longitude ?? 0);

    var countryNameList = placemarks[0].country?.split(' ');

    if (countryNameList!.isNotEmpty && countryNameList.length >= 2) {
      for (var eachLetter in countryNameList) {
        abbr += eachLetter[0];
      }
    } else {
      abbr = countryNameList.toString().substring(0, 2).toUpperCase();
    }
    return UserLocation(
        city: placemarks[0].locality ?? 'Chennai',
        country: abbr,
        latitude: _currentPosition.latitude,
        longitude: _currentPosition.longitude);
  }

现在，当我调用这个函数时，它说它返回 Future

无论如何，关键是每当我从其他地方调用此函数时，我都想访问此方法返回的 UserLocation 对象，但它总是说此函数返回 Future。我怎么能这样做？有什么想法吗？

Answer 1

这应该可以解决问题，将getUserLocation()函数的返回类型设置为Future<UserLocation?>，并在没有任何返回的地方返回null。