javascript - How do I reduce multiple Promise.all?

Question

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, input.orgId),
      EmailQueries.addEmail(user.emails, user.userId, input.orgId),
      PhoneQueries.addPhones(user.phones, user.userId, input.orgId)
    ])
    .then(() => Promise.resolve(user))
  }, Promise);
})

How could I perform this operation?

There's an obvious case typo between `promise` and `Promise`, anyway. — Tatsuyuki Ishi, Apr 16 '17 at 04:31
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, Apr 16 '17 at 04:36
There is no `Promise.prototype.all()`. Only `Promise.all()`. — Tatsuyuki Ishi, Apr 16 '17 at 04:38
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, Apr 16 '17 at 04:39
Obviously you don't understand. Move on to the next question. — jhamm, Apr 16 '17 at 04:42
@jhamm - no, you don't understand - there is no `Promise.prototype.all` - your code as written will fail with errors — Jaromanda X, Apr 16 '17 at 05:43

score 3 · Accepted Answer · answered Apr 16 '17 at 04:46

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, input.orgId),
      EmailQueries.addEmail(user.emails, user.userId, input.orgId),
      PhoneQueries.addPhones(user.phones, user.userId, input.orgId)
    ]))
    .then(() => user))
  }, Promise.resolve());
})

I tried this, but I only get the last user in the result. I don't get the rest of the users that are inserted. — jhamm, Apr 16 '17 at 04:53
What do you mean by "I don't get the rest of the users that are inserted."? There is no need to use `.reduce()` if expected result is array of results. What is purpose for using `.reduce()`? — guest271314, Apr 16 '17 at 04:54

score 2 · Answer 2 · answered Apr 16 '17 at 04:48

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, input.orgId),
      EmailQueries.addEmail(user.emails, user.userId, input.orgId),
      PhoneQueries.addPhones(user.phones, user.userId, input.orgId)
    ]);
  }));
});

javascript - How do I reduce multiple Promise.all?

2 Answers2

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