如何按顺序执行承诺，从数组传递参数？

Question

var myArray = [1, 2, 3, 4, 5, 6]

function myPromise(num){
  return new Promise(res => {
    window.setTimeout(()=>{
      res(  console.log("done: " + num)  )
    },2000)
  })
}


myPromise(myArray[0])
  .then(x => myPromise(myArray[1]))
  .then(x => myPromise(myArray[2]))
  .then(x => myPromise(myArray[3]))
  .then(x => myPromise(myArray[4]))
  .then(x => myPromise(myArray[5]))

Right now, if I execute the statement above, it will run sequentially. In my actual use case the array is dynamically populated and I need to execute the myPromise() function for each member in myArray .

How can I make a "pauseable loop" that will loop for each item in the array, execute myPromise and wait for the promise to be resolved before continuing to the next iteration?

Answer 1

You can make the repeated application of .then into a fold pretty neatly if you're okay with creating as many promises as array elements as is the case in the question:

myArray.reduce(
  (p, x) =>
    p.then(_ => myPromise(x)),
  Promise.resolve()
)

but an async function, for example, doesn't require that:

const mapSeries = async (iterable, action) => {
  for (const x of iterable) {
    await action(x)
  }
}

mapSeries(myArray, myPromise)

which is built into the excellent promise library Bluebird as mapSeries :

Promise.mapSeries(myArray, myPromise)

Runnable snippet:

 const myArray = [1, 2, 3, 4, 5, 6] const sleep = ms => new Promise(res => { setTimeout(res, ms) }) const myPromise = num => sleep(500).then(() => { console.log('done: ' + num) }) myArray.reduce( (p, x) => p.then(_ => myPromise(x)), Promise.resolve() )

Answer 2

Don't create an array of promises. Create an array of functions returning a promise.

const f = x => new Promise(resolve => setTimeout(() => resolve(console.log(x)), 2000))

(async () => {
    for (let job of [1, 2, 3, 4, 5, 6].map(x => () => f(x)))
        await job()
})()

Promises start running immediately after creation. Therefore, sequential execution is ensured by constructing the next promise only after finishing the current one.

Answer 3

Also you can do it via recursive approach - executeSequentially calls itself:

 function createPromise(x) { return new Promise(res => { setTimeout(() => { console.log(x) res(x); }, x * 1000) }) } function executeSequentially(array) { return createPromise(array.shift()) .then(x => array.length == 0 ? x : executeSequentially(array)); } console.time('executeSequentially'); executeSequentially([1, 2, 3]).then(x => { console.log('last value: ' + x); console.timeEnd('executeSequentially'); });

Answer 4

Sequential:

you can use async await features to run promises sequentially . here's a snippet

async function chainPromiseCalls(asyncFunctions=[],respectiveParams=[]){

    for(let i=0;i<asyncFunctions.length;i++){
        const eachResult = await asyncFunctions[i](...respectiveParams[i]);
        // do what you want to do with each result 
       
    }

    return ;
}

Parallel:

for parallel you can just call each async function once in a loop , but if you do want to get their combined result , you can use Promise.all

function parallelPromiseCalls(asyncFunctions=[],respectiveParams=[]){
    return Promise.all(asyncFunctions.map((func,index)=>func(...respectiveParams[index])))
       .then(resultsList=>{
        resultsList.forEach((result,index)=>{
           // do what you want to do with each result in the list
        })
        return ;
    })
}

note : I am considering respective parameters as an list of lists since multiple parameters should be passed to any one of the function , else if you have to pass only a single parameter to each then you can remove the spread operator.

Answer 5

I know I am very late, and my answer is similar to what others have posted. But I thought I could post a clearer answer which may help any beginner.

Instead of using promises directly, we can use promise factory. Since the promise starts executing as soon as they are created using promise factory we delay the creation of promise.

In this example I create 5 which resolve after a second. I use a promiseCreator to create promises. Now the array promises uses promiseCreator to create 5 instances of promises. But array promiseFactories wraps promiseCreator in a function so promise is not invoked immediately. It is invoked when used.

Function executeSequentially executes all promiseLike sequentially.

• When promise array is passed result is promise array itself executes parallely (actually they executed as soon as they are created, not when this line is called).
• When promiseFactory array is passed result is new Promise is created when earlier promise has completed their execution.

 const promiseCreator = (i, time, text) => { return new Promise(resolve => setTimeout( () => resolve(console.log(`${i} ${text}`)), time) ); } const promises = [ promiseCreator(1, 1000, "parallel"), promiseCreator(2, 1000, "parallel"), promiseCreator(3, 1000, "parallel"), promiseCreator(4, 1000, "parallel"), promiseCreator(5, 1000, "parallel"), ] const promiseFactories = [ () => promiseCreator(1, 1000, "sequential"), () => promiseCreator(2, 1000, "sequential"), () => promiseCreator(3, 1000, "sequential"), () => promiseCreator(4, 1000, "sequential"), () => promiseCreator(5, 1000, "sequential"), ] function executeSequentially(promiseLikeArray) { var result = Promise.resolve(); promiseLikeArray.forEach(function (promiseLike) { result = result.then(promiseLike); }); return result; } executeSequentially(promises) executeSequentially(promiseFactories)

Answer 6

You could use Array.reduce .

//type: [number]
var myArray = [1, 2, 3, 4, 5, 6] //doesn't really matter

//type: number -> Promise<number>
function myPromise(num){
  return new Promise((resolve) => {
    window.setTimeout(()=>{
      resolve(console.log("done: " + num)  )
    },2000)
  })
}

//Array.reduce has type: [a] ~> ((b, a) -> b), b) -> b
//So it can have type:
//[number] ~> ((Promise<number>, number) -> Promise<number>), Promise<number>) -> Promise<number>
//Therefore we need to give reduce a function that takes a Promise 
//resolving to a number and a number which makes a new promise.
//This is the function we want:

function sequencePromises(promise, number) {
  return new Promise((resolve) => {
    resolve(promise.then(_ => myPromise(number)));
  });
} 

myArray.reduce(sequencePromises, Promise.resolve());

Of course, this simplistic approach won't work if you have a promise which can error, or if you need previous results, so you might want to make sequencePromises more generic:

function genericSequencePromises(promiseFunction) {
  return (promise, parameter) => {
    return new Promise((resolve, reject) => 
                         return promiseFunction(resolve, 
                                                reject, 
                                                promise, 
                                                parameter));
  }
}

Then you can do whatever you want as long as you return a Promise.

Finally, you might benefit from this little helper:

function promiseSeries(array, reducer) {
  return array.reduce(reducer, Promise.resolve());
}

Bringing it all together:

let sequencePromises = genericSequencePromises((resolve, reject, promise, num) => {
  resolve(promise.then(_ => console.log(`done: ${num}`)));
}

promiseSeries(myArray, sequencePromises);

This way, you can not only handle the case in your question, but much more complex cases.

Answer 7

I would use babel and do it this way:

 let args = [1, 2, 3]; const myPromise = async x => console.log('arg:',x); const test = async () => { for (let task of args.map(myPromise)) await task; } test().then(console.log('Done'));

 <script src="https://unpkg.com/babel-standalone@6.24.0/babel.min.js"></script>

Answer 8

You can iterate over the array of elements and pass the params like this:

 const arr = [1, 2, 3, 4, 5, 6]; const MyPromiseFunction = num => new Promise((resolve, reject) => { // Your logic... setTimeout(() => num <= 4 ? resolve('Success!') : reject('Rejected!'), 1000 * num); }); const logMessage = (num, msg) => console.log(`For number ${num} promise result: ${msg}`); arr.forEach( num => MyPromiseFunction(num) .then(message => logMessage(num, message)) .catch(reason => logMessage(num, reason)) );

Answer 9

 const myArray = [1, 2, 1000, 4, 5000, 6] const sleep = ms => new Promise(res => { setTimeout(res, ms) }) const myPromise = num => sleep(num).then(() => { console.log('done: ' + num) }) myArray.reduce( (p, x) => p.then(_ => myPromise(x)), Promise.resolve() )