android - 函数的返回类型为 'Future'，但没有以return语句flutter结尾

Question

我正在尝试从flutter中的函数返回Future对象，我必须在数据库查询块中添加return语句，无法从外部访问它，但是我遇到了这个错误Function has a return type of 'Future<Station>' but doesn't end with a return statement flutter，因为它有望找到返回值函数末尾的语句，如果完整的示例在这里:

  Future<Station> gettingStationData(String stationName) async {
    var databasesPath = await getDatabasesPath();
    String path = join(databasesPath, 'main.db');
    Database database = await openDatabase(path, version: 1);
    database.transaction((tnx) async {
      List<Map> stationQuery = await tnx.rawQuery(
          "SELECT * FROM Station WHERE stationName = ?", [stationName]);
      Station station = Station(
          key: stationQuery[0]['key'],
          stationName: stationQuery[0]['stationName'],
          cityName: stationQuery[0]['cityName'],
          stationLatitude: stationQuery[0]['stationLatitude'],
          stationLongitude: stationQuery[0]['stationLongitude']);
      return station;
    });
  }

每当我尝试在函数末尾添加 return station;时，都会给我 unresolved reference错误。每当我尝试访问该函数时，它都会返回 null。有什么帮助吗？

Answer 1

您需要返回database.transaction(...)返回的内容。

return database.transcation((tnx) async { ...