Question

我有3种类型：

export interface Animal {
...
}

export interface Cat extends Animal{
...
}

export interface Dog extends Animal{
...
}

export interface Iguana extends Animal {
...
}

在一个单独的文件中，我使用以下语法定义了3个异步函数：

const task1 = (() => Promise<Cat[]>) = async () => {
...
}

const task2 = (() => Promise<Dog[]>) = async () => {
...
}

const task3 = (() => Promise<Iguana[]>) = async () => {
...
}

const tasks = [task1, task2, task3];
const runningTasks = tasks.map(task => task());
// below line throws an error
const results = await Promise.all(runningTasks);

由于类型不兼容，最后一行引发错误，并且这非常长，并且基本上说“狗缺少Cat的属性”。

我想做的就是简单地异步调用这三个函数并保存它们的结果。

Answer 1

我想做的就是简单地异步调用这三个函数并保存它们的结果。

您具有这种设置：

export interface Animal {
    name: string;
}

export interface Cat extends Animal {
    attitude: string;
}

export interface Dog extends Animal {
    sleepDuration: number;
}

export interface Iguana extends Animal {
    skinToughness: number;
}

const task1 = async (): Promise<Cat[]> => {
    return await Promise.resolve([]);
}

const task2 = async (): Promise<Dog[]> => {
    return await Promise.resolve([]);
}

const task3 = async (): Promise<Iguana[]> => {
    return await Promise.resolve([]);
}

一种方法是像这样使用Promise中的父类型：

const demo = async () => {
    const tasks = [task1, task2, task3];
    const runningTasks: Promise<Animal[]>[] = tasks.map(task => task());
    const results: Animal[][] = await Promise.all(runningTasks);
}

如果从结果中使用特定类型很重要，则jcalz建议的元组将起作用：

const demo = async () => {
    const tasks = [task1, task2, task3] as const;

    type TaskResults = [
        ReturnType<typeof task1>,
        ReturnType<typeof task2>,
        ReturnType<typeof task3>,
    ];

    const runningTasks = tasks.map(task => task()) as TaskResults;
    const results = await Promise.all(runningTasks);

    results[0][0].attitude;
    results[1][0].sleepDuration;
    results[2][0].skinToughness;
}

令人惊讶的是，如果您不需要提前启动任务，则可以使用这种方法维护类型信息：

const demo = async () => {

    const results = await Promise.all([
        task1(),
        task2(),
        task3()
    ])

    results[0][0].attitude;
    results[1][0].sleepDuration;
    results[2][0].skinToughness;
}

Promise.all在调用具有不同返回类型的多个异步函数时抛出类型错误

1 个答案: