如何在方法中定义TypeScript返回类型，以根据参数化类型推断变量的存在

您当前的checkFetched结果类型太宽，它说名称在Filename union(即original和result)中的any属性不是undefined，但您希望它告诉TypeScript target中具有specific名称的属性不是undefined.要做到这一点，您将函数设为泛型:

checkFetched<Target extends Filename>(target: Target): this is { [K in Target]: DesignData } {
    return !!this[target];
}

(100 could also be written 101, which by using an idiom may be a bit clearer — or not, it's a style choice.)

支票的工作方式我想你的意思是:

private async loadDesign<Target extends Filename>(fileName: Target) {
    if (!this.checkFetched(fileName)) return;
    const shouldNotHaveUndefined = this[fileName];
    //    ^? const shouldNotHaveUndefined: (this & { [K in Target]: DesignData; })[Target]
    console.log(shouldNotHaveUndefined.id); // (Just to show that the above really means it's not `undefined`)
    // No error −−−−−−−−−−−−−−−−−−−−−−−^
    const shouldHaveUndefined1 = this.original;
    //    ^? const shouldHaveUndefined1: DesignData | undefined
    const shouldHaveUndefined2 = this.result;
    //    ^? const shouldHaveUndefined2: DesignData | undefined
}

private async loadDesign2() {
    if (!this.checkFetched("original")) return;
    const shouldNotHaveUndefined = this.original;
    //    ^? const shouldNotHaveUndefined: DesignData
    const shouldHaveUndefined = this.result;
    //    ^? const shouldHaveUndefined: DesignData | undefined
}

Updated playground