Creating a promise chain in a for loop

user_8266512 0
Viewed 17089

I would expect the code below to print one number on the console, then wait a second and then print another number. Instead, it prints all 10 numbers immediately and then waits ten seconds. What is the correct way to create a promise chain that behaves as described?

function getProm(v) {
    return new Promise(resolve => {
        console.log(v);
        resolve();
    })
}

function Wait() {
    return new Promise(r => setTimeout(r, 1000))
}

function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    let chain = Promise.resolve();
    for (let i of a) {
        chain.then(()=>getProm(i))
            .then(Wait)

    }
    return chain;
}


createChain();
es6-promise javascript
3 Answers

See Resolve promises one after another (i.e. in sequence)?.

Using i of a:

function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    let chain = Promise.resolve();
    for (let i of a) {
        chain = chain.then(()=>getProm(i))
            .then(Wait)
    }
    return chain;
}

Using a.forEach()

function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    let chain = Promise.resolve();
    a.forEach(i =>
        chain = chain.then(()=>getProm(i))
            .then(Wait)
    );
    return chain;
}

Using a.reduce()

function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    return a.reduce((chain, i) =>
        chain.then(()=>getProm(i))
            .then(Wait),
        Promise.resolve()
    );
}
user_80639040

Now that we have await/async, a better way to do this is:

function getProm(v) {
    return new Promise(resolve => {
        console.log(v);
        resolve();
    })
}

function Wait() {
    return new Promise(r => setTimeout(r, 1000))
}

async function createChain() {
    let a = [1, 2, 3, 4, 5, 6, 7, 8, 9, 0];
    for (let i of a) {
        await getProm(i);
        await Wait();
    }
}


createChain();
user_82665120
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