在JavaScript中循环通过一系列承诺

Question

我有一些代码可以通过一系列承诺循环，并输出值。

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const promises = [
  wait(1),
  wait(3),
  wait(2),
  wait(4),
  wait(5),
];

for (var promise of promises) {
  promise.then(seconds => console.log(`waited ${seconds} seconds`));
}

这样的问题是，承诺结果不会以数组的顺序记录。我的预期结果是：

Waited 1 seconds
Waited 3 seconds
Waited 2 seconds
Waited 4 seconds
Waited 5 seconds

结果是：

Waited 1 seconds
Waited 2 seconds
Waited 3 seconds
Waited 4 seconds
Waited 5 seconds

所以我想拥有这样的事情，

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const promises = [
  wait(1),
  wait(3),
  wait(2),
  wait(4),
  wait(5),
];

for (var promise of promises) {
  // When the promise is resolved, log `Waited ${seconds} seconds`
}

我该怎么做？

Answer 1

如果您不希望计时器一次开始全部启动，则不应一次进行所有这些等待一次调用。相反，只有在上一个产生的承诺得到解决时，只有下一个电话：

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const tasks = [
  () => wait(1),
  () => wait(3),
  () => wait(2),
  () => wait(4),
  () => wait(5),
];

(async function () {
  for (const task of tasks) {
    const seconds = await task();
    console.log(`waited ${seconds} seconds more`);
  }
})();

如果您 do 想要同时开始所有承诺，并且想要等待它们的全部在开始输出结果之前解决，然后使用Promise> Promise .all：

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const promises = [
  wait(1),
  wait(3),
  wait(2),
  wait(4),
  wait(5),
];

Promise.all(promises).then(values => {
  console.log("all results are in:");
  for (const seconds of values) {
    console.log(`waited ${seconds} seconds (from the start)`);
  }
});

如果您 do 希望所有承诺要同时开始，并且想等待它们仅在上一个解决方案后才解决，请使用等待：

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const promises = [
  wait(1),
  wait(3),
  wait(2),
  wait(4),
  wait(5),
];

(async function () {
  for (const promise of promises) {
    const seconds = await promise;
    console.log(`waited ${seconds} seconds (from the start)`);
  }
})();

Answer 2

丹尼尔·怀特（Daniel A.

function wait(seconds) {
  return new Promise((resolve, reject) => {
    setTimeout(() => {
      resolve(seconds);
    }, seconds * 1000);
  });
}

const promises = [wait(1), wait(3), wait(2), wait(4), wait(5)];

Promise.all(promises).then((results) => {
  for (var seconds of results) {
    console.log(`waited ${seconds} seconds`);
  }
});