JavaScript Array Comparison Function

Question

I currently have a map of an array of users which all have a unique _id key / value.

user = [{_id: "1", ... }, {_id: "2", ... }, ... ]

I also have two other arrays, one named teams and another named accounts.

teams = [{ _id: "1", members: [{ userId: "2" }, { userId: "4" }, ... ], ... }]

accounts = [{ _id: "1", authorizedUsers: [{ userId: "3"}, ... ], ownerTeamId: "2" }, ... ]

Trying to create two comparison functions which takes the argument of user and outputs numberOfTeams and numberOfAccounts for the corresponding user.

I have attempted the numberOfTeams below but I'm not sure if it's the most optimal.

numberOfTeams(user) {
  let count = 0;
  teams.forEach(team => {
    team.members.forEach(member => {
      if (member.userId === user._id) {
        count++
      }
    })
  });
  return count;
}

With the numberOfAccounts, I'm stuck on how to compare authorizedUsers === user._id OR ownerTeamId === team._id where also members.userId === user.id, and then count++.

Answer 1

It’s probably a good start to write a function to get the teams a user belongs to:

function containsUserId(users, id) {
    return users.some(user => user.userId === id);
}

function getUserTeams(user, teams) {
    return teams.filter(team =>
        containsUserId(team.members, user._id));
}

because then you can write numberOfTeams using it:

numberOfTeams(user) {
    return getUserTeams(user, teams).length;
}

then a similar function to get accounts:

function getUserAccounts(user, accounts) {
    const userTeamIds = new Set(
        getUserTeams(user).map(team => team._id)
    );

    return accounts.filter(account =>
        containsUserId(account.authorizedUsers, user._id) ||
        userTeamIds.has(accounts.ownerTeamId));
}

then numberOfAccounts using it:

numberOfAccounts(user) {
    return getUserAccounts(user, accounts).length;
}

Essentially: use more functions so you can understand the steps you’re taking to solve your own problem and, in doing so, use those steps more effectively.