ts 如何根据可选参数返回不同的泛型成员，有简单的写法吗？

[a132811]

我想实现一个 ts 版本的简版 usePromise

  // 期望类型推导出：data 的类型是 {list: FileNode[]} (即返回 T)
  const [data, isOk] = usePromise(
    () => {
      return Promise.resolve({list:[]}) as Promise<{ list: FileNode[] }>;
    },
    { initValue: {list:[]} }
  );

  // 期望类型推导出：fileNodes 的类型是一个 FileNode 数组 (即返回 T['list'])
  const [fileNodes, isOk] = usePromise(
    () => {
      return Promise.resolve({list:[]}) as Promise<{ list: FileNode[] }>;
    },
    { initValue: [] },
    "list",
  );
  

我的 usePromise 实现如下, 利用K extends keyof T 判断如果 K 属性存在就返回T[K], 否则返回T 测试发现 ts 判断可选参数时总会失败，只会返回T.

export function usePromise<T, K>(
  factory: () => Promise<T>,
  options: Options<K extends keyof T ? T[K] : T> = {},
  filterKey?: K,
): [K extends keyof T ? T[K] : T, boolean] {
  type R = K extends keyof T ? T[K] : T;
  const [state, setState] = useState<R>(
    options.initValue!,
  );
  const isLoadingRef = useRef(false);
  useEffect(() => {
    factory().then((r) => {
      if (filterKey) {
        setState((r as any)[filterKey] as unknown as R);
      } else {
        setState(r as unknown as R);
      }
    }).catch((res) => {
      if (options.onError) options.onError(res);
      else throw res;
    });
  }, []);
  return [state, isLoadingRef.current] as [R, boolean];
}

我知道可以增加两个 function 的类型重载解决这个问题，但是手写一遍类型重载感觉太麻烦。

请教一下有没有更简单的方法呢？

[YuJianrong]

为什么一定要纠结范型呢，你这个例子用 overloading 做起来好多了：

https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgGLADYQHIHsAmEA3gL4BQoksiKA8gA5jC4gDOAPACoB8RyowMADU4GAK4QA-AC5knZOTIIWrMMgDa+OGDgAafq1oBrAIIBdZAF5kY1hAAKUXAFtgdgBRlkydwEor3MhEXt7IUBBgYlAgyI4ubhAAdOGsuBgAbhDuRBhuYNLqZiT+cKyxTq527Hy5qrLoWHiEhQrcANwhJLohfALCohKyOXkFRQpkvh1kAPTTyID45oDY5oDePoDR6oAVxoA-2oBfioBY-zCYOAQQZYAhbmuA9GaAAHKAVHJoh00ogA6mgCN+PoDOyoAr8YB7anLqAOS1MD-My+JQqNTqA6NY6sfRuYwAIQs1lsDgqCU83j8ASCIW84Ui0XK8TsyROaUy2SBo2KyFKJMqEGqyBp9xhzQsJHanW63l6IEEInEEFkLS6IQAREDJd1JmQFRAAB70XBQNQwMQgBBMFg2OxxJlcfQAaWQysgIHwZSMEAAnrgYHJuO54Dq1XbZDjLIFDQkuNx9LhGMw2LIGLq2Fx1CazIHkNCqCb7bITb4xZwY2Z9AAjXBpCBwEAWDrK1XqhNanWh-Xo0nMniuxBgD1e-w+xn+nhBkMqcO9qM8dMaTi5-NYIslshltUaquR2t+qqj5Bmi0QK02+2O51YhPN1s+du+jHLwMhYOR1j9q-R2PIAA+zqsQQl3kT0GTdpkq7lGazj5yGOBaTniBIRFEMQgGIGAYPSZRFnaHQkEAA

[YuJianrong]

虽然范型也不是不行：
https://www.typescriptlang.org/play?#code/JYOwLgpgTgZghgYwgAgGLADYQHIHsAmEA3gL4BQoksiKA8gA5jC4gDOAPACoB8RyowMADU4GAK4QA-AC5knZOTIIWrMMgDa+OGDgAafq1oBrAIIBdZAF5kY1hAAKUXAFtgdgBRlkydwEor3MhEXt7IUBBgYlAgyI4ubhAAdOGsuBgAbhDuRBhuYNLqZiT+cKyxTq527Hy5qrLoWHiEhQrcANwhJLohfALCohKyOXkFRQpkvh1kAPTTyID45oDY5oDePoDR6oAVxoA-2oBfioBY-zCYOAQQZYAhbmuA9GaAAHKAVHJoh00ogA6mgCN+PoDOyoAr8YB7anLqAOS1MD-My+JQqNTqA6NY6sfRuYwAIQs1lsDgqCU83j8ASCIW84Ui0XK8TsyROaUy2SBo2KyFKJMqEGqyBp9xhzQsJHanW63l6IEEInEEFkLS6IQAREDJd1JmQyBAAB70XBQNQwMQgBBMFg2OxxJlcfQAaWQysgIHwZSMEAAnrgYHJkAAfGxWiAHEAQfBWd2EL0+7hY5DwHVqu2yHGWQKGhJcbh85C4RjMNiyBi6tjsM0WiBWm32x3OyR-E0WWQ8P2kJPQqgm+0yZAmuVi3NKy3W5C2h1O+SlzjqcvISv6ABGuDSEDgIAswQJESiMRAYgwGHpZRndo6JCAA

不过真不如 overloading 。

[a132811]

@YuJianrong 不是纠结范型。只是想找一个比 overloaded 更简单的方法。

在 statckoverflow 我找到答案啦：
https://stackoverflow.com/questions/73718892/return-different-generic-type-via-functions-optional-argument-in-typescript