javascript await on multiple chained async functions

Question

Say I have the following:

const a = new A();
await a.getB().action();

A.prototype.getB() is async as-well as B.prototype.action(). If I try to await on the chaining of the functions I get an error: TypeError: a.getB(...).action is not a function.

If I am separating the chaining of the functions and awaiting each promise it works fine. Is there a way to chain these promises and await them together?

Try: `await (await a.getB()).action()` or `await a.getB().then(result => result.action())` — Titus, Jul 07 '19 at 12:54
I was hoping there's some syntatic sugar I could get away with, guess I will have to await for each promise — Jorayen, Jul 07 '19 at 13:04

PVermeer · Answer 1 · 2019-07-07T13:08:48.173

4

You need to await hem both:

const a = new A();
const b = await a.getB();
await b.action();

Or

const a = new A();
await a.getB().then(b => b.action());

edited Jul 07 '19 at 13:08

answered Jul 07 '19 at 12:54

PVermeer

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`var` is a keyword, and it will throw a error when you try to run this code `variable name is missing` – Code Maniac Jul 07 '19 at 13:07

score 3 · Accepted Answer · answered Jul 07 '19 at 13:02

3

This is because getB is an async function and does not return a B object but a Promise object that have no action method. This promise will further be resolved with a B object and you can access to the resolved value by catching it with the then method as proposed by PVermeer.

answered Jul 07 '19 at 13:02

F. Bauer

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Thanks for clarifying that. I was just about to make an edit :) – PVermeer Jul 07 '19 at 13:06

javascript await on multiple chained async functions

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