类型 'boolean' 不可分配给类型 'U[T]'答案

【问题标题】：Type 'boolean' is not assignable to type 'U[T]'类型 'boolean' 不可分配给类型 'U[T]'
【发布时间】：2021-11-16 17:08:04
【问题描述】：

我不明白为什么 typescript 给我这个错误：

这是相关代码：

  getFeaturesFromMembership$<T extends keyof MembershipFeatures,
    U extends Record<T, boolean>>(membership$: Observable<Membership>,
                                  featureKeys: T[],
                                  isPreview$: Observable<boolean> = of(false)
  ): Observable<U> {
    return combineLatest([membership$, isPreview$])
      .pipe(
        map(([membership, isPreview]) => {
          const features: U = {} as U;

          for (const key of featureKeys) {
            features[key] = false; // <- error here
          }

          if (!membership || isPreview) {
            Object.keys(features).forEach(key => features[key] = true);
            return features;
          }

          Object.keys(features).forEach(key => features[key] = key in membership.features ? membership.features[key] : true);

          return features;
        })
      );
  }

Playground example

【问题讨论】：

请提供可重现的例子
我添加了一个游乐场示例链接
这样不行。关于U，请看这个答案stackoverflow.com/questions/69187022/…。有用吗？
另一方面，对象的突变可能会导致您出现另一种类型的错误。请参阅我关于 TS 中的突变的文章catchts.com/mutations
见这个例子 tsplay.dev/mx51GW 。 U 可能是Record<T, boolean> 的任何允许子类型

标签： typescript

【解决方案1】：

这样的事情怎么样...请注意，我重构了 for 循环。

// I did create this types for simulating your types
type Membership = { features : string[] }
type Observable<T> = any
function of<T>(someValue : T) : Observable<T>{}
interface MembershipFeatures { key1: string; key2: string; }


// My solution
type FoundRecord<T extends keyof MembershipFeatures> = { [key in T] : boolean}

// Function...
function getFeaturesFromMembership$<SelectedKey extends keyof MembershipFeatures>(
    membership$: Observable<Membership>,
    featureKeys: SelectedKey[],
    isPreview$: Observable<boolean> = of(false)
  ): Observable<FoundRecord<SelectedKey>> {
    return combineLatest([membership$, isPreview$])
      .pipe(
        map(([membership, isPreview]) => {
            const features: FoundRecord<SelectedKey> = {} as FoundRecord<SelectedKey>;
            for (const key of featureKeys) {
                if (!membership || isPreview) {
                    features[key] = true;
                } else {
                    features[key] = false;
                    features[key] = key in membership.features ? membership.features[key] : true
                }
            }

            return features;
        })
      );
  }

【讨论】：