I am trying to use a Promise.all inside of a reduce and cannot get my function to work, unless there is only one user in my array. The starting object of the reduce is a Promise. The first time through the reduce, the Promise has .all available on it. The second time through, the .all is not available.

return UserQueries.addUsersOnCasefileCreate(input).then(users => {
  return users.reduce((promise, user) => {
    return promise.all([
      AddressQueries.addAddress(user.address, user.userId, inputId),
      EmailQueries.addEmail(user.emails, user.userId, inputId),
      PhoneQueries.addPhones(user.phones, user.userId, inputId)
    ])
    .then(() => Promise.resolve(user))
  }, Promise);
})

How could I perform this operation?

I am trying to use a Promise.all inside of a reduce and cannot get my function to work, unless there is only one user in my array. The starting object of the reduce is a Promise. The first time through the reduce, the Promise has .all available on it. The second time through, the .all is not available.

return UserQueries.addUsersOnCasefileCreate(input).then(users => {
  return users.reduce((promise, user) => {
    return promise.all([
      AddressQueries.addAddress(user.address, user.userId, inputId),
      EmailQueries.addEmail(user.emails, user.userId, inputId),
      PhoneQueries.addPhones(user.phones, user.userId, inputId)
    ])
    .then(() => Promise.resolve(user))
  }, Promise);
})

How could I perform this operation?

• javascript
• promise
• reduce

• There's an obvious case typo between promise and Promise, anyway. – Tatsuyuki Ishi Commented Apr 16, 2017 at 4:31
• The promise that is lower case is just the first argument of the reduce referring to the Promise as the initial object. It isn't a typo. – jhamm Commented Apr 16, 2017 at 4:36
• There is no Promise.prototype.all(). Only Promise.all(). – Tatsuyuki Ishi Commented Apr 16, 2017 at 4:38
• The promise from the first argument is the Promise in the initial object. That is why it works the first time through the loop. That is why Promise.all is available at first. – jhamm Commented Apr 16, 2017 at 4:39
• 1 @jhamm - no, you don't understand - there is no Promise.prototype.all - your code as written will fail with errors – Jaromanda X Commented Apr 16, 2017 at 5:43

2 Answers 2

You initialize with Promise which is a function, though return a resolved Promise object, where the two are not the same.

You can initialize with Promise.resolve(), call promise.then(), then return Promise.all() with .then() chained within first .then(), which passes Promise object to next iteration at .reduce().

return UserQueries.addUsersOnCasefileCreate(input).then(users => {
  return users.reduce((promise, user) => {
    return promise.then(() => Promise.all([
      AddressQueries.addAddress(user.address, user.userId, inputId),
      EmailQueries.addEmail(user.emails, user.userId, inputId),
      PhoneQueries.addPhones(user.phones, user.userId, inputId)
    ]))
    .then(() => user))
  }, Promise.resolve());
})

There's no need to use reduce(). Just map the things and wait them all.

return UserQueries.addUsersOnCasefileCreate(input).then(users => {
  return Promise.all(users.map((user) => {
    return Promise.all([
      AddressQueries.addAddress(user.address, user.userId, inputId),
      EmailQueries.addEmail(user.emails, user.userId, inputId),
      PhoneQueries.addPhones(user.phones, user.userId, inputId)
    ]);
  }));
});