Question 51098 in Javascript Question posted in
A very good W3school tutorial can be found here.

How rewrite this code using promise chaining? – Javascript

function getSmth(num) {
  return new Promise((resolve, reject) => {
    setTimeout(() => resolve(num * num), 500)
  });
}
function func() {
  getSmth(2).then(res1 => {
    getSmth(3).then(res2 => {
      getSmth(4).then(res3 => {
        console.log(res1 + res2 + res3);
      });
    });
  });
}
   func();
function getSmth(num) {
  return new Promise((resolve, reject) => {
    setTimeout(() => resolve(num * num), 500)
  });
}
function func() {
  getSmth(2)
    .then(res1 => getSmth(3))
    .then(res2 => getSmth(4))
    .then(res3 => {
      console.log(res1 + res2 + res3);
    })
}
func();

Should output 29 to the console but it’s not work.

Tags:

2

Answers


  1. 0

    As others have mentioned before me the issue with your chaining is that you are losing the context of the previous result. So you can either use async/await or Promise.all

    Here is the async/await syntax which leaves the code more readable

    function getSmth(num) {
  return new Promise((resolve, reject) => {
    setTimeout(() => resolve(num * num), 500)
  });
}

async func(){
  const res1 = await getSmth(2);
  const res2 = await getSmth(3); 
  const res3 = await getSmth(4);
  console.log(res1 + res2 +res3);
}

func().then(()=>{})

    You can also use Promise.all like this

    
Promise.all([getSmth(2), getSmth(3), getSmth(4)])
.then( results => console.log(results[0] + results[1] + results[2]);
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  2. 0

    If you want to use async/await other answers show the correct code. I think what you might be trying to get is:

    function func() {
  getSmth(2)
    .then(result => getSmth(3).then(res2 => result + res2))
    .then(result => getSmth(4).then(res3 => result + res3))
    .then(result => {
      console.log(result);
    })
}
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