CODEWITHSUNDEEP
node.jsasynchronouspromise
Savas P
Savas P

Reputation: 95

Properly construct return value using node.js Promises

I am new to node.js and I am having hard time with Promises. I want to construct/build my result variable step by step by using Promises. I "abstracted" my code just to point out the issue better. Basically what I am trying to do is to create a resulting model of several rest api calls (those calls that can be done in parallel are called in Promise.all).

Thank you in advance.

function test() {
    var result = {}; 
    var prom1 = new Promise(function(resolve, reject) {
        resolve(addTwo(result)); 
    }).catch(err => console.log(err));
    return prom1.then(function(result) {
            promises = [];
            promises.push(new Promise(function(resolve, reject) {
                resolve(addD(result)); 
            }));
            promises.push(new Promise(function(resolve, reject) {
                resolve(addC(result));
            }));
            Promise.all(promises)
                         .then(result) 
    }).then(console.log(result)); //logging because I was testing
}

function addTwo(result) {
    result.a = "a";
    result.b = "b";
    return result;
}

function addD(result) {
    result.d = "d";
}

function addC(result) {
    result.c = "c";
}

test();

The output that expected was: { a: 'a', b: 'b', d: 'd', c: 'c' }, but instead I got: { a: 'a', b: 'b' }

I understand that if I call then() on a Promise, that I will have in that block access to the return value from the promise, but can I adjust my code somehow to "build" the result variable in the then calls with using Promise.all at some point?

Upvotes: 0

Views: 67

Answers (2)

alx.lzt
alx.lzt

Reputation: 486

In a more ES8 way with the async/await syntax you could escape the .then().catch() struggle. I feel like it's easier to get how promises work this way.

async function test() {
    try {
        let result = {}
        await new Promise(function(resolve, reject) {
            resolve(addTwo(result))
        })
        let promises = []
        promises.push(new Promise(function(resolve, reject) {
            resolve(addD(result))
        }))
        promises.push(new Promise(function(resolve, reject) {
            resolve(addC(result))
        }))
        await Promise.all(promises)
        return result
    } catch (e) {
        console.error(e)
    }
}

Note that you'll have to await your test function result

Upvotes: 0

GProst
GProst

Reputation: 10227

  1. You need to return your Promise.all(promises) so that its result is chained to the then where you have console.log(result).
  2. I believe you have an error at the line Promise.all(promises).then(result), you pass result to then but then expects a function as an argument, not an object

Consider using async/await cause it's less confusing than these Promise chains

Upvotes: 1

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