Question

更新了问题：

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。所有这些子类都有自己的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。在这个主处理程序中，我想使用type来决定调用哪个特定的处理程序。

由于事件是Event子类的实例，因此我使用event.constructor.type 它编译并正常工作，但IDE抱怨：Property 'type' does not exist on type 'Function'（我在代码示例中也标记了它）。

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));

当这段代码编译并运行时（控制台输出“SubClass1”），编辑器会抱怨：“函数”类型中不存在属性“类型”。

我在Intellij Idea和Typescript游乐场（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构造函数创建一个类型：

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

然后投向它：

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

从实例

1 个答案: