使用Typescript中的数组迭代来简化async-await

问题描述：

有没有一种简单的方法可以在Typescript中表达这种语法，而无需Promise.all和Array.prototype.map?

Is there a simpler way to express this syntax in Typescript, without Promise.all and Array.prototype.map?

const items = [...];
await Promise.all(items.map(async item => {
    doSomething(item);
    const result = await doSomethingAsync(item);
    doSomethingMore(result, item);
});

答

ES5数组方法不完全支持async和生成器函数，这就是为什么for和其他循环语句应优先于forEach在ES6中.

ES5 array methods don't fully support async and generator functions, that's one of the reasons why for and other loop statements should be preferred to forEach in ES6.

在这种情况下，map被部分滥用，因为它不管理数组值.如果应并行解决承诺，则可能应该是:

In this case map is partially misused, because it doesn't manage array values. If promises should be resolved in parallel, it likely should be:

items = await Promise.all(items.map(async item => {
    doSomething(item);
    const result = await doSomethingAsync(item);
    doSomethingMore(result, item);
    return item;
});

没有那么简单，但在语义上可能是正确的.如果do..函数将item修改为其类型更改的点，则const newItems = await Promise.all(...)也可以更轻松地管理项目类型.

Not much simpler but possibly semantically correct. If do.. functions modify item to the point its type changes, const newItems = await Promise.all(...) also makes it easier to manage item types.

如果应该依次解决这些问题，而不必涉及Promise.all，则可以为for..of:

If they should be resolved in series and no Promise.all has to be involved, this can be for..of:

for (const item of items) {
    doSomething(item);
    const result = await doSomethingAsync(item);
    doSomethingMore(result, item);
});

使用Typescript中的数组迭代来简化async-await

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