cant return the chained promises result

Question

const request = require('request-promise')

required this module and use it in this way the data and subData is options that i defined later...

    const foo= (data, subData) => {
    return request(data)
     .then(result => {
        console.log('result:',result)
        return request(subData)
    })
}

the problem is the request(data) result is not return but the request(subData) result is return

Q.allSettled([
  foo(),
  fo(),
  f(),
  .
  .
  .
])

and with q module create an array of promises, but i still cant get my expected return result

because you are not returning the response from `request(data)` from `foo`, just remove your `foo` method and use `Q.allSettled([request(data), request(subData)])` — AZ_, Oct 14 '19 at 08:32
@AZ_ i need to call request(subData) after the request(data) was finished. — khodekazemi, Oct 14 '19 at 10:04
then add another `.then` block and return an array of both responses. — AZ_, Oct 14 '19 at 10:06
can u show me the syntax iam new to promise concept and nodejs @AZ_ — khodekazemi, Oct 14 '19 at 10:09

score 1 · Accepted Answer · answered Oct 14 '19 at 10:19

You can use any of the following methods to chain your promises and return both responses into an array

const foo = (data, subData) => {
    let result;
    return request(data)
        .then(res => {
            result = res;
            return request(subData)
        }).then(res => {
            return [result, res]
        });
}

//OR
const foo2 = (data, subData) => {
    return request(data)
        .then(res1 => {
            return request(subData).then(res2 => {
                return [res1, res2]
            })
        });
}

//OR
const foo3 = async (data, subData) => {
    let res1  = await request(data);
    let res2 = await request(subData);
    return [res1, re2];
}

score 0 · Answer 2 · answered Oct 14 '19 at 08:11

0

I would suggest the following approach instead of Promise Chaining or Q:

const fetchData = (data, subData) => {    
    const result = await request(data);
    const resultSub = await request(subData);

    console.log(result, resultSub);
};

answered Oct 14 '19 at 08:11

Horatiu Jeflea

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cant return the chained promises result

2 Answers2