参数类型“FutureOr<void> Function(Database, int)”无法分配给参数类型“FutureOr<void> Function(Database, int)？”

Question

这个 [ int = int? ] 我知道这是一个错误，
但是 [ FutureOr func() = FutureOr func()？ ]
我不明白这之间的区别
func() 和 func()？
请有人向我解释一下这是怎么回事吗？

这是有问题的代码：

        // create a getter to instantiate the database obj
         Future<Database> get database async {
        // initialize the database when uninitialized
         if (_database != null) return _database!;
           _database = await _initDB('students.db');
           return _database!;
         }
       Future<Database> _initDB(String filePath) async {
         final dbPath = await getDatabasesPath();
         final path = join(dbPath, filePath);
         return await openDatabase(path, version: 1, onCreate: _createDB);
       }
      FutureOr<void> _createDB(Database db, int version) async {
        const String idType = 'INTEGER PRIMARY KEY AUTOINCREMENT';
        const String textType = 'TEXT NOT NULL';
        const String boolType = 'BOOLEAN NOT NULL';
        const String intType = 'INTEGER NOT NULL';
        const String telephoneType = 'INTEGER NOT NULL';
    
        const String studentID = 'StudentId';
        const String studentFirstName = 'studentFirstName';
        const String studentMiddleName = 'studentMiddleName';
        const String studentLastName = 'studentLastName';
        const String studentAge = 'studentAge';
        const String studentGender = 'studentGender';
        const String studentHometown = 'studentHometown';
    const String studentLocation = 'studentLocation';


    return await db.execute('''
      CREATE TABLE  Students(
        $studentID $idType,
        $studentFirstName $textType,
        $studentMiddleName $textType,
        $studentLastName $textType,
        $studentAge $intType,
        $studentGender $textType,
        $studentHometown $textType,
        $studentLocation $textType,
      )
      ''');
    }  

这也是为了更好的语法突出显示的图像

Answer 1

我相信你必须将这些返回类型更改为可为空

Database?

Future<Database?> get database async {
        // initialize the database when uninitialized
         if (_database != null) return _database!;
           _database = await _initDB('students.db');
           return _database!;
         }
       Future<Database?> _initDB(String filePath) async {
         final dbPath = await getDatabasesPath();
         final path = join(dbPath, filePath);
         return await openDatabase(path, version: 1, onCreate: _createDB);
       }

Answer 2

我相信你必须更改 _createdDB，它的返回类型是 Future 并且你返回 db.execute 函数。

FutureOr<void> _createDB(Database db, int 
version) async {
    

等待 db.execute(''' 创建表学生（ $studentID $idType, $studentFirstName $textType, $studentMiddleName $textType, $studentLastName $textType, $studentAge $intType, $studentGender $textType, $studentHometown $textType, $studentLocation $textType， ） ''');