如何在到达下一行之前等待 Promise.all() 完成？

Question

我正在学习 Node.js。

我必须在循环work()内调用异步函数Promise.all()，并且必须在继续执行 Promise.all() 之后的语句之前完成该函数。目前，它到达了FINISH未完成的状态work()。

让代码等待 work() 函数完成的正确方法是什么？

const promise1 = Promise.resolve(1);
const promise2 = Promise.resolve(2);
const promise3 = Promise.resolve(3);

async function work() {
    await new Promise((resolve, reject) => {
        setTimeout(resolve, 2000, 'foo');
    })
    console.log('some work here')
}

async function main() {
    await Promise.all([promise1, promise2, promise3]).then((values) => {
        values.forEach(function(item) {
            console.log(item)
            work()
        });
    });
    console.log('FINISH')
}

main()

Answer 1

很难说你在这里真正想要什么，但一般来说不要混合await搭配。then

const promise1 = Promise.resolve(1);
const promise2 = Promise.resolve(2);
const promise3 = Promise.resolve(3);

function delay(ms) {
  return new Promise((resolve) => setTimeout(resolve, ms));
}

async function work(value) {
  await delay(1000 * value);
  console.log("some work here", value);
  await delay(1000);
  return value * 2;
}

async function main() {
  // Doesn't really do much, since these are already resolved...
  const values = await Promise.all([promise1, promise2, promise3]);
  // Pass all of those values to `work`, which returns a promise,
  // and wait for all of those promises to resolve.
  const workedValues = await Promise.all(values.map(work));
  console.log(workedValues);
}

main();

打印出来（第一行有各种延迟）

some work here 1
some work here 2
some work here 3
[ 2, 4, 6 ]