Question

鉴于this code sample in TypeScript playground，我无法正确键入greeterBuilderName变量，以便可以将其传递给getProp实用程序... keyof Greeter实际上是道具/方法的列表Greeter的实例，如何指定我想要Greeter func本身的道具（prototype，warm_greeter，...）？

换句话说，greeterBuilderName的正确类型是什么？

谢谢！

function getProp<T, K extends keyof T>(object: T, property: K): T[K] {
    return object[property];
}

class Greeter {
    private greeting: string;
    public constructor(message: string) {
        this.greeting = message;
    }
    public greet(): string {
        return "Hello, " + this.greeting;
    }
    public static warm_greeter(): Greeter { return new Greeter('good o\' fellow'); }
}

let greeterBuilder: () => Greeter = getProp(Greeter, 'warm_greeter'); // ok
alert(greeterBuilder().greet()); // ok

const greeterBuilderName: keyof Greeter = 'warm_greeter'; // does not build :(
greeterBuilder = getProp(Greeter, greeterBuilderName); // does not build :(
alert(greeterBuilder().greet());

Answer 1

我想你想获得Greeter类本身的键，所以

const greeterBuilderName: keyof typeof Greeter = 'warm_greeter';

typeof类的静态属性的keyof

1 个答案: