Возможно ли обработать Promise.all без второго Promise.all?

Question

[AlexNew22]

Возможно ли обработать Promise.all без второго Promise.all?
https://jsfiddle.net/7ts2rcgw/

Comments

[Alexandroppolus]

у меня jsfiddle не открывается, кинь сюда код

[AlexNew22]

Alexandroppolus,

const urls = [
  'https://api.github.com/users/iliakan',
  'https://api.github.com/users/remy',
  'https://api.github.com/users/jeresig'
]

const requests = urls.map(item => fetch(item))

Promise.all(requests).then(data => Promise.all(data.map(item => item.json()))).then(data => data.forEach(item => console.log(item.id)))

Answer 1

[0xD34F]

Promise
  .all(urls.map(url => fetch(url).then(result => result.json())))
  .then(results => results.forEach(result => console.log(result.id)));

Answer 2

[zkrvndm]

Конечно можно:

const urls = [
  'https://api.github.com/users/iliakan',
  'https://api.github.com/users/remy',
  'https://api.github.com/users/jeresig'
]

const requests = urls.map(async item => (await fetch(item)).json());

Promise.all(requests).then(responses => console.log(responses));