如果每个promise已经有一个catch块，则抛出Promise.all（[]）catch块？

Question

I thought that this would throw an error in the Promise.all catch block, but it never gets that far?

I'm trying to understand how I can handle the rejected promise at the call and also at Promise.all .

const apiCallOne = new Promise((resolve, reject) => (
  resolve('Resolved !!!')
)).then(console.log)
  .catch(console.warn);

const apiCallTwo = new Promise((resolve, reject) => (
  reject('Rejected !!!')
)).then(console.log)
  .catch(console.warn);

Promise.all([apiCallOne, apiCallTwo])
  .then(value => console.log('all', value))
  .catch(err => console.error('error', err));

Will Promise.all ever hit it's catch block?

Answer 1

No it won't the catch will swallow the errors and like you said Promise.all catch block will never be called, the following will work according to your expectations

const apiCallOne = new Promise((resolve, reject) => (
  resolve('Resolved !!!')
)).then(console.log)
  .catch((err) => {
    console.warn(err)
    throw err
  });

const apiCallTwo = new Promise((resolve, reject) => (
  reject('Rejected !!!')
)).then(console.log)
  .catch((err) => {
    console.warn(err)
    throw err
  });

Promise.all([apiCallOne, apiCallTwo])
  .then(value => console.log('all', value))
  .catch(err => console.error('error', err));

Answer 2

You can handle them when you call Promise.all , there is no need to handle them before you execute the requests because we want to run them in parallel and cancel all if there is an error.

const apiCallOne = new Promise((resolve, reject) => (
  resolve('Resolved !!!')
))

const apiCallTwo = new Promise((resolve, reject) => (
  reject('Rejected !!!')
))

Promise.all([apiCallOne, apiCallTwo])
  .then(value => console.log('all', value))
  .catch(err => console.error('error', err));