合并两个对象数组，跳过具有相同ID属性的对象

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. , 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"
    }

];

Answer 1

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); 

Answer 2

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); 

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Answer 3

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];

Answer 4

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

Answer 5

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

2. Merge both arrays.

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

Answer 6

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

    for(var i=0; i<localUsers.length; ++i) {
        var checked = 0;
        for(var j=i+1; j<apiUsers.length; ++j) {
             if(localUsers[i]['id'] === apiUsers[j]['id']){
                  apiUsers.splice(j--, 1);
             }
        }
    }
    console.log(localUsers.concat(apiUsers));