typescript - 从实例获取类(静态)属性

Question

更新的问题:

在客户端应用程序中，我正在实现一个事件处理系统，类似于 Redux 的工作方式。
我将事件类型定义为自定义 Event 类的子类。所有此类子类都有自己的 type 静态属性。
我通过创建子类的实例并将其发送到主处理函数来触发事件。 在此主处理程序中，我想使用 type 来决定调用哪个特定处理程序。

由于事件是 Event 子类的实例，因此我使用 event.constructor.type。
它编译并工作正常，但 IDE 提示:类型“Function”上不存在属性“type”(我也在代码示例中标记了它)。

我应该忽略该消息，还是有更好的方法从实例访问该静态属性？

type State = {};

class Event {
    public static type: string;
}

class MouseMoveEvent extends Event {
    public static type = "MouseMoveEvent";
    constructor(public x: number, public y: number) {
        super();
    }
}

class KeypressEvent extends Event {
    public static type = "KeyPressEvent";
    constructor(public key: string) {
        super();
    }
}

function handleMouseMove(event: MouseMoveEvent, state: State): State {
    // return some new state
}

function handleKeyPress(event: KeyPressEvent, state: State): State {
    // return some new state
}

const handlerMap: { [type: string]: (event: Event, state: State) => State; } = {
    [MouseMoveEvent.type]: (event: Event, state: State) => handleMouseMove(event as MouseMoveEvent, state),
    [KeyPressEvent.type]: (event: Event, state: State) => handleKeyPress(event as KeyPressEvent, state)
    // etc.
};

// The main handler, it receives all triggered events, and decides which specific handler to call, based on the `type` property.
function handleEvent(event: Event, state: State): State {
    // the editor complains here
    return handlerMap[event.constructor.type](event, state);
}

原始问题:

class BaseClass {
    public static type: string;
}

class SubClass1 extends BaseClass {
    public static type = "SubClass1";
}

class SubClass2 extends BaseClass {
    public static type = "SubClass2";
}

const map: {[type: string]: number} = {
    [SubClass1.type]: 1,
    [SubClass2.type]: 2
};

function getNumber(x: BaseClass): number {
    return map[x.constructor.type]; // complains here
}

const foo = new SubClass1();
console.log(getNumber(foo));

当此代码编译并运行时(控制台输出“SubClass1”)，编辑器提示:“Function”类型上不存在属性“type”。

我在 Intellij Idea 和 Typescript Playground (www.typescriptlang.org/play) 中尝试过。我应该忽略编辑器消息，还是有更好的方法从实例访问该静态属性？

Answer 1

您可以为 BaseClass 构造函数创建类型:

interface BaseClassStatic {
    type: string;
    new (): BaseClass;
}

然后转换到它:

function getNumber(x: BaseClass): number {
    return map[(x.constructor as BaseClassStatic).type];
}