Question

public class Base{
    protected Base instantiate() {//Should return an instance of the subclass that this method has been called from


    }
}

public class Sub1 extends Base {

}

public class Sub2 extends Base {

}

public static void main(String[] a) {
    Sub1 sub1 = new Sub1();
    Sub2 sub2 = new Sub2();
    Sub1 ss = sub1.instantiate();//should return a new instance of Sub1 
    Sub2 s = sub2.instantiate();//should return a new instance of Sub2
}

如何在超类中实例化子类。换句话说，我需要知道上面代码中Base类中的instantiate（）方法的实现。

修改：这是实际代码的一部分：基类是：

     /**
         * Any class that extends this class should override the TABLE_NAME
         *
         * @author sSh
         */
        public class Model {

            protected static final String TABLE_NAME = null;

            private void createRecord() throws ClassNotFoundException, SQLException {


            }

            private void updateRecord() {  

            }

            public ResultSet getAllRecords() { 

                //This is the method--that returns instance of the subclass that this method has 
   // been called from
            }

            public void saveRecord() throws SQLException, ClassNotFoundException {

            }


        }

Model类应该是所有模型的Base类。所以，如果我们有PatientModel，它将是这样的：

public class PatientModel extends Model{

        protected static final TABLE_NAME = "patient";

        //...fields and methods


    }


    public static void main(String[] a){
        PatientModel.getAllRecords();//this will return all the records in patient table as patient objects

    }

Answer 1

你不能这样做，因为你不能覆盖静态方法。

您可以做的是创建一个执行此任务的BaseService类。

片段：

public class BaseService {
      ....
      public static Base instantiate(Class<? extends Base> clazz) {
          //create the new instance based upon the parameter
      }
}

在main中：

 Base base = BaseService.instantiate(Sub1.class);

或者如果你想将Sub1实例作为Sub1类型的变量，你应该强制转换。

如何在超类中实例化子类？

1 个答案: