对Array.map使用async await [英] Use async await with Array.map

问题描述

给出以下代码：

var arr = [1,2,3,4,5];

var results: number[] = await arr.map(async (item): Promise<number> => {
        await callAsynchronousOperation(item);
        return item + 1;
    });

产生以下错误：

TS2322：类型'Promise< number> []'不能分配给'number []'。
输入'Promise< number>不能指定为'number'类型。

TS2322: Type 'Promise<number>[]' is not assignable to type 'number[]'. Type 'Promise<number> is not assignable to type 'number'.

我该如何解决？如何让异步等待和 Array.map 一起工作？

How can I fix it? How can I make async await and Array.map work together?

推荐答案

这里的问题是你试图等待一系列承诺而不是承诺。这不符合您的预期。

The problem here is that you are trying to await an array of promises rather than a promise. This doesn't do what you expect.

当传递给的对象await 不是Promise时， await 只是立即返回值，而不是尝试解决它。所以既然你传递了 await 一个数组（Promise对象）而不是Promise，await返回的值就是那个类型为承诺< number> [] 。

When the object passed to await is not a Promise, await simply returns the value as-is immediately instead of trying to resolve it. So since you passed await an array (of Promise objects) here instead of a Promise, the value returned by await is simply that array, which is of type Promise<number>[].

您需要做的是致电 Promise.all 在 map 返回的数组上，以便在 await 之前将其转换为单个Promise。

What you need to do here is call Promise.all on the array returned by map in order to convert it to a single Promise before awaiting it.

根据 MDN文档 Promise.all ：

Promise.all（iterable）方法返回一个promise，当可迭代参数中的所有promise都已解析时，它会解析
，或
因为第一次通过承诺拒绝而拒绝。

The Promise.all(iterable) method returns a promise that resolves when all of the promises in the iterable argument have resolved, or rejects with the reason of the first passed promise that rejects.

所以在你的情况下：

var arr = [1, 2, 3, 4, 5];

var results: number[] = await Promise.all(arr.map(async (item): Promise<number> => {
    await callAsynchronousOperation(item);
    return item + 1;
}));

这将解决您在此遇到的具体错误。

This will resolve the specific error you are encountering here.

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