Return values from SetTimeout function ( Promise.Then())

Question

In the code below , Values are RETURNED correctly from a queued Promise.then() chain .

CODE:

let cond_1 = true;
let data = 'Data Received....';
let err  = 'Error';

var p1 = new Promise(function(resolve,reject){ 
    if(cond_1){
        resolve(data);
    }else{
        reject(err); }})

p1.then((data)=>{console.log(data);return 'Wait....';})
.then((val1)=>{console.log(val1); return 'Finished';})
.then((val2)=>{console.log(val2)})
.catch((err)=>{console.log(err)});

Output :

Data Received....
Wait....
Finished

However, the same RETURNED values from a chained SetTimeout function are returned 'UNDEFINED'.

CODE:

p1.then((data)=>{console.log(data); return 'Wait.....'; })
.then((val1)=>{setTimeout(function(val1){console.log(val1); return 'Finished';},1000)})
.then((val2)=>{setTimeout(function(val2){console.log(val2);},1000)})
.catch((err)=>{console.log(err)});

Output:

Data Received....
undefined
undefined

How to resolve this?

`setTimeout` does not return a promise. Just a timer ID. And you don't return that one either, in your code. "Solving" this depends on what you actually want to do. You could [make a delay](https://stackoverflow.com/a/39914235) or even modify it slightly to produce a value after some time. But I'm not really sure what the intention is here. Why would you need to use `setTimeout`? — VLAZ, Mar 19 '22 at 10:38
@VLAZ ... The above script is just a short test for queued execution with PRE_DEFINED DELAY and the code in the function block would like to replaced a larger codeblock. However the code EXECUTES AT ONCE (SetTimeout@1000ms) instead of each block executing after 1000ms ONE AFTER ANOTHER. ..... Are you saying that a new PROMISE has to be created for each .Then() ? — srt111, Mar 19 '22 at 12:31

score 1 · Answer 1 · answered Mar 19 '22 at 10:54

1

Try taking advantage of Lexicographic nature of Javascript.

Instead of making a function v1,v2 which your functions takes within setTimeout, just use an arrow function. In this way you are using the v1,v2 returned from promise.

Do this

let cond_1 = true;
let data = 'Data Received....';
let err = 'Error';

var p1 = new Promise(function(resolve, reject) {
  if (cond_1) {
    resolve(data);
  } else {
    reject(err);
  }
})

p1.then((data) => {
    console.log(data);
    return 'Wait.....';
  })
  .then((val1) => {
    setTimeout(() => {
      console.log(val1);
    }, 1000);
    return 'Finished';
  })
  .then((val2) => {
    return setTimeout(() => {
      console.log(val2)
    }, 1000)
  })
  .catch((err) => {
    console.log(err)
  });

What you did was you created a new variable v1,v2 for your function. You can only use that when you pass value v1,v2 in that function. That function won't use v1,v2 returned from promise as you expect.

answered Mar 19 '22 at 10:54

Shivansh Verma

138
6

"*Lexicographic nature of Javascript.*" do you mean [lexical](https://en.wikipedia.org/wiki/Scope_(computer_science)#Lexical_scope) instead of [lexicographic](https://en.wikipedia.org/wiki/Lexicographic_order)? – VLAZ Mar 19 '22 at 11:00
@Shivansh Verma ....The code EXECUTES AT ONCE (SetTimeout@1000ms) instead of each block executing after 1000ms - 'ONE AFTER ANOTHER'. – srt111 Mar 19 '22 at 12:32
@VLAZ ,@Shivansh Verma ...Any way to chain the SetTimeout with a delay of 1000ms 'EACH' (rather than each SetTimeout@1000ms getting executed at once). – srt111 Mar 19 '22 at 13:02
@VLAZ yes I mean lexical nature. You are right. – Shivansh Verma Mar 21 '22 at 06:53
@starzar If you mean to execute a block after some definite interval again n again lets say 1000ms then you should use setInterval instead of setTimeout – Shivansh Verma Mar 21 '22 at 06:55

Return values from SetTimeout function ( Promise.Then())

1 Answers1