Promise.all 不適用於異步函數數組

Question

理論上，考慮以下代碼，它在 I/O 操作完成后將消息打印到控制台。

const foo = (num) => new Promise(resolve => setTimeout(resolve, num * 1000)); // An async I/O function in actual code
array = [[1, 2, 3], [1, 2, 3] , [1, 2, 3]];

const promiseArray = array.map(arr => {
  arr.map(num => {
    return (async () => {
      await foo(num);
      console.log(num);
    });
  });
}).flat();

await Promise.all(promiseArray);

我不知道為什么，但它不起作用。 控制台上沒有打印任何內容。

---

但是，如果我將異步 function 包裝在 Promise 構造函數中，它將起作用

const foo = (num) => new Promise(resolve => setTimeout(resolve, num * 1000)); // An async I/O function in actual code
array = [[1, 2, 3], [1, 2, 3] , [1, 2, 3]];

const promiseArray = array.map(arr => {
  arr.map(num => {
    return new Promise(async () => {
      await foo(num);
      console.log(num);
    });
  });
}).flat();

await Promise.all(promiseArray);

我應該如何重寫代碼以擺脫 Promise 構造函數？

Answer 1

Promise.all將一組承諾作為其參數，而不是一組async function s。 此外，您還缺少return聲明。 你應該寫

const promiseArray = array.flatMap(arr => {
  return arr.map(async num => {
    await foo(num);
    console.log(num);
  });
});

await Promise.all(promiseArray);

或者

const promiseArray = array.map(async arr => {
  await Promise.all(arr.map(async num => {
    await foo(num);
    console.log(num);
  }));
});

await Promise.all(promiseArray);

Answer 2

Its normal Promise.all take an array of Promises, async function are of type function, but returns a Promise once invoked if no explicite return it will return a resolved promise with undefined value.

 async function myAsyncFunction(){ return 1; } console.log(typeof myAsyncFunction) console.log(typeof myAsyncFunction()) console.log(myAsyncFunction() instanceof Promise)

Answer 3

您從 map 回調返回 function，而不是 promise。 而是返回foo(num) 。 然后在展平之后你有一系列的承諾。

const foo = (num) => new Promise(resolve => setTimeout(resolve, num * 1000)); // An async I/O function in actual code
array = [[1, 2, 3], [1, 2, 3] , [1, 2, 3]];

const promiseArray = array.map(arr => {
  return arr.map(foo); // its equal arr.map(num => foo(num));
}).flat();

const results = await Promise.all(promiseArray);
results.forEach(item => console.log(item));

Answer 4

異步function 必須返回 promise。 您的第一個實現需要一個 return 語句：

const array = [[1, 2, 3], [1, 2, 3] , [1, 2, 3]];

const promiseArray = array.map(arr => {
  // returning the mapped list
  return arr.map(async (num) => {
    const result = await foo(num);
    console.log('Num: ' + num);
    // return at the end of async will generate a promise fulfillment for you.
    // see: https://developers.google.com/web/fundamentals/primers/async-functions
    return result;
  });
}).flat();

const result = await Promise.all(promiseArray);
console.log('result: ' + result);