Question

我想合并两个对象数组，但是我想跳过具有相同ID的对象（我只想保存具有相同ID的第一个对象）。

一个数组存储在本地，另一个数组从API获取用户。

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"
    }
];

我希望得到这个。 ID为2的apiUsers中的对象被跳过，因为该对象已经存在于localUsers对象数组中。我想对所有具有相同ID的对象执行此操作。

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

创建您的mergedUsers来连接localUser和已经不在localUser中的apiUser：

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

您可以通过按所需顺序减少数组来使用Map。

Lon

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);

Answer 3

首先从apiUsers中删除localUsers中存在的所有实例，然后将该数组添加到localUsers中。数组的顺序与问题中没有提到的无关，但很容易执行。

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

Answer 4

您可以使用它来将两个对象与重复项组合在一起：

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

然后，您可以使用新对象删除重复项。

引用：Merge 2 arrays of objects

Answer 5

从apiUserArray中过滤出已经在localUserArray中的用户。

合并两个数组。

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

let mergedArray = localUsers.concat(distinctApiUsers)

Answer 6

编写自己的比较功能很容易：


    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));

合并两个对象数组，并跳过具有相同ID属性的对象

6 个答案: