How to return the Promise.all fetch api json data?

I came across the same problem and my goal was to make the Promise.all() return an array of JSON objects.

let jsonArray = await Promise.all(requests.map(url => fetch(url)))
    .then(async (res) => {
        return Promise.all(
            res.map(async (data) => await data.json())
        )
    })

And if you need it very short (one liner for the copy paste people :)

const requests = ['myapi.com/list','myapi.com/trending']
const x = await Promise.all(requests.map(url=>fetch(url))).then(async(res)=>Promise.all(res.map(async(data)=>await data.json())))

Instead of return [aa.json(),bb.json()] use return Promise.resolve([aa.json(),bb.json()]). See documentation on Promise.resolve() .

Aparently aa.json() and bb.json() are returned before being resolved, adding async/await to that will solve the problem :

.then(async([aa, bb]) => {
    const a = await aa.json();
    const b = await bb.json();
    return [a, b]
  })

Promise.all([
    fetch('https://jsonplaceholder.typicode.com/todos/1'),
    fetch('https://jsonplaceholder.typicode.com/todos/2')
  ]).then(async([aa, bb]) => {
    const a = await aa.json();
    const b = await bb.json();
    return [a, b]
  })
  .then((responseText) => {
    console.log(responseText);

  }).catch((err) => {
    console.log(err);
  });

Still looking for a better explaination though

The simplest solution would be to repeat the use of Promise.all, to just wait for all .json() to resolve. Just use :

Promise.all([fetch1, ... fetchX])
.then(results => Promise.all(results.map(r => r.json())) )
.then(results => { You have results[0, ..., X] available as objects })