Merge two array of objects with skipping objects that has same ID property

Question

I wanna merge two arrays of objects but I want to skip the objects that has the same ID (i want to save only first object that has same id).

One array is stored locally and the other I'm fetching users from API.

const localUsers = [
    {
        "id": 1,
        "first_name": "Adam",
        "last_name": "Bent",
        "avatar": "some img url"
    },
    {
        "id": 2,
        "first_name": "New Name",
        "last_name": "New Last Name",
        "avatar": "some new img url"
    }

];

const apiUsers = [
    {
        "id": 2,
        "first_name": "Eve",
        "last_name": "Holt",
        "avatar": "some img url"
    },
    {
        "id": 3,
        "first_name": "Charles",
        "last_name": "Morris",
        "avatar": "some img url"
    }
];

I expect to get this. The object in apiUsers with the id: 2 is skipped, because he already exist in the localUsers array of objects. I want to do this to all the objects with the same id.

const mergedUsers = [
    {
        "id": 1,
        "first_name": "Adam",
        "last_name": "Bent",
        "avatar": "some img url"
    },
    {
        "id": 2,
        "first_name": "New Name",
        "last_name": "New Last Name",
        "avatar": "some new img url"
    },
    {
        "id": 3,
        "first_name": "Charles",
        "last_name": "Morris",
        "avatar": "some img url"
    }

];

user_2652759 · Answer

Create your mergedUsers concatenating localUsers and the apiUsers that are not in localUsers already:

const localUsers = [
    {
        "id": 1,
        "first_name": "Adam",
        "last_name": "Bent",
        "avatar": "some img url"
    },
    {
        "id": 2,
        "first_name": "New Name",
        "last_name": "New Last Name",
        "avatar": "some new img url"
    }

];

const apiUsers = [
    {
        "id": 2,
        "first_name": "Eve",
        "last_name": "Holt",
        "avatar": "some img url"
    },
    {
        "id": 3,
        "first_name": "Charles",
        "last_name": "Morris",
        "avatar": "some img url"
    }
];

const mergedUsers = localUsers.concat(apiUsers.filter(a => !localUsers.find(b => b.id === a.id)));
console.log(mergedUsers);

user_1447675 · Answer

You could take a Map by reducing the arrays in the wanted order.

const
    localUsers = [{ id: 1, first_name: "Adam", last_name: "Bent", avatar: "some img url" }, { id: 2, first_name: "New Name", last_name: "New Last Name", avatar: "some new img url" }],
    apiUsers = [{ id: 2, first_name: "Eve", last_name: "Holt", avatar: "some img url" }, { id: 3, first_name: "Charles", last_name: "Morris", avatar: "some img url" }],
    result = Array.from(
        [localUsers, apiUsers]
            .reduce(
                (m, a) => a.reduce((n, o) => n.set(o.id, n.get(o.id) || o), m),
                new Map
            )
            .values()
    );

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

user_1898688 · Answer

First remove all instance from apiUsers that exist in localUsers, then add that array to localUsers. The order of the array does not matter here as not stated in the question, but easy to perform.

const filteredApiUsers = apiUsers.filter(x => !localUsers.some(y => x.id === y.id));
const mergedUsers = [...localUsers, ...filteredApiUsers];

user_9118924 · Answer

you can use this to combine the two objects with duplicates:

Array.prototype.push.apply(localUsers,apiUsers);

then you can remove duplicates with the new object.

reference : Merge 2 arrays of objects

user_5547718 · Answer

Filter out users from apiUserArray that are already in the localUserArray.

Merge both arrays.

let distinctApiUsers = apiUsers
                        .filter(rUser => localUsers
                                          .findIndex(lUser => lUser.id == rUser.id) == -1)

let mergedArray = localUsers.concat(distinctApiUsers)

user_8526380 · Answer

It's easy enough to write your own comparison function:


    for(var i=0; i

Merge two array of objects with skipping objects that has same ID property

6 Answers