javascript - 合并两个对象数组并跳过具有相同 ID 属性的对象

Question

我想合并两个对象数组，但我想跳过具有相同 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，另一个我从 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。 apiUsers中id为2的对象被跳过，因为他已经存在于localUsers对象数组中。我想对具有相同 ID 的所有对象执行此操作。

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

创建你的 mergedUsers 连接 localUsers 和已经不在 localUsers 中的 apiUsers:

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);