Question

不知道这是怎么回事。我对我的类型做了奇怪的事情吗？

const reflect = (promise): Promise<Reflection> =>
    promise.then(
        (value) => ({ value, resolved: true }),
        (error) => ({ error, rejected: true })
    );

const to = (promiseArr) => {
    return Promise.all(promiseArr.map(reflect)).then((sources: Reflection[]) => sources);
};

Argument of type '(sources: Reflection[]) => Reflection[]' is not assignable to parameter of type '(value: [unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown]) => Reflection[] | PromiseLike<Reflection[]>'.
  Types of parameters 'sources' and 'value' are incompatible.
    Type '[unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown, unknown]' is not assignable to type 'Reflection[]'.
      Type 'unknown' is not assignable to type 'Reflection'.ts(2345)

Answer 1

这取决于您的Reflection类型，但是如果Promise解析，听起来好像应该使用一个resolve值的类型参数，

type Reflection<T> = {
    value: T;
    resolved: true;
} | {
    error: unknown;
    rejected: true;
}

然后，在reflect和to中，确保表示参数的类型，并将这些类型作为类型参数传递：

const reflect = <T>(promise: Promise<T>): Promise<Reflection<T>> =>
    promise.then(
        value => ({ value, resolved: true }),
        error => ({ error, rejected: true })
    );

const to = <T>(promiseArr: Array<Promise<T>>) => {
    return Promise.all(promiseArr.map(reflect)).then((sources: Array<Reflection<T>>) => sources);
};

这可以正确编译，并且TS将to的类型检测为：

const to: <T>(promiseArr: Promise<T>[]) => Promise<Reflection<T>[]>

不过，请注意，.then中的最后一个to并没有做任何事情，因此您可以将其简化为

const to = <T>(promiseArr: Array<Promise<T>>) => {
    return Promise.all(promiseArr.map(reflect))
};

Working demo

如何为Promise.all定义返回类型？

1 个答案: