javascript - Node 异步/等待与 promise.all

Question

在一个简单的 JS 程序中，我需要让 asyncOperation2 和 asyncOperation3 与 asyncOperation1 一起顺序执行。这意味着我需要发生以下情况:

1) order of execution as  1,2,3 or 2,3,1
2) then after step 1. I need to preform an operation on the result of 1,2,3

这是我目前所拥有的(这将在 Node 中运行)。但是我无法按照上面描述的那样获得操作顺序。

const App = {
    total: 0,

    init: function () {
        // I want to run all 3 operations. (with 2 and 3 happening sequentially)
        Promise.all([App.asyncOperation1(1000), App.combine2and3()])
            .then((total) => {
                // then I want to use the result of all 3 operations
                console.log(`total from top: ${total}`);
            })
    },

    asyncOperation1: function (time) {
        return new Promise(function (resolve, reject) {
            var intervalID = setTimeout(function () {
                resolve(time);
            }, time);
        });
    },

    asyncOperation2: async function (time) {
        setTimeout(function () {
            return time;
        }, time);
    },

    asyncOperation3: async function (time) {
        setTimeout(function () {
            return time;
        }, time);
    },

    combine2and3: async function () {
        var value2 = await App.asyncOperation2(2000);
        var value3 = await App.asyncOperation3(1000);
        console.log(`value2: ${value2}`)
        console.log(`value3: ${value3}`)
        console.log(`summed total ${value2 + value3}`);
        return value2 + value3;
    },
};

App.init();

实际结果:

value2: undefined
value3: undefined
summed total NaN
total from top: 1000,NaN

期望的结果:

value2: 2000
value3: 1000
summed total 3000
total from top: 1000,3000

Answer 1

问题是 asyncOperation2()和 asyncOperation3()还一个 promise (因为它们的 async 声明)没有解析值(因此它是 undefined )因为这些函数没有自己的返回值。

asyncOperation2: async function (time) {
    setTimeout(function () {
        return time;     // this just returns back into the timer sub-system
    }, time);
    // there is no return value for your function here
    // thus the promise the async function returns has an undefined resolved value
},

从 setTimeout() 内部返回不是函数的返回值。这只是在函数本身返回很久之后返回到定时器内部。记住，setTimeout()是非阻塞的。它安排一些代码在将来的某个时间运行，然后立即返回，然后您的函数完成并返回(在 setTimeout() 触发之前)。那么，稍后，您的 setTimeout()触发并且您从该回调返回一个值，该值仅返回到计时器子系统，而不返回到您的任何代码或任何 promise 。

将这两个函数改成这样:

function delay(t, v) {
   return new Promise(resolve => {
       setTimeout(resolve.bind(null, v));
   }, t);
}

asyncOperation2: function (time) {
    return delay(time, time);
},

asyncOperation3: function (time) {
    return delay(time, time);
},

现在您有函数返回一个用期望值解决的 promise 。请注意，不需要声明它们 async要么因为你没有使用 await在函数内部，您已经创建了自己的返回 promise 。