设计模式在Java中建模动作

Question

我希望使用设计模式来标准化常见操作，但我不确定哪一个是最好的

假设我们从两个服务Java类开始，每个类有两个操作/方法

class Service1 {
    public void performSomething() {
        // Some complex algorithm implemented here
    }
    public void performSomethingElse {
        // Some complex algorithm implemented here
    }
}

class Service2 {
    public void performSomething() {
        // Some complex algorithm implemented here
    }
    public void performSomethingElse {
        // Some complex algorithm implemented here
    }
}

两个服务共享相同的算法，所以很自然，我想重构performSomething()和performSomethingElse()。我的方法是为每个重构方法创建两个单一的方法类

interface Action {
    public void run();
}
class PerformSomething implements Action {
    public void run() {}
}
class PerformSomethingElse implements Action {
    public void run() {}
}

class Service1 {
    private PerformSomething algo1;
    private PerformSomethingElse algo2;

    public void businessUseCase1() {
        algo1.run();
        algo2.run();
    }

}

我觉得这种简单的方法很幼稚，我非常确定有一种更合适的设计模式可以表示一个动作，而不是创建一个自定义接口来表示动作

Answer 1

考虑到亲吻和春天，我会做以下几件事：

@Service
class Service1 {
    public void performSomething() {
        // Some complex algorithm implemented here
    }
    public void performSomethingElse {
        // Some complex algorithm implemented here
    }
    public void businessUseCase1() {
        performSomething();
        performSomethingElse();
    }

}

@Service
class Service2 {
    @Autowired
    Service1 service1;

    public void performSomething() {
        // Some complex algorithm implemented here
    }
    public void performSomethingElse {
        // Some complex algorithm implemented here
    }
    public void businessUseCase1() {
        service1.performSomething();
        service1.performSomethingElse();
        // or
        service1.businessUseCase1();
    }

}