如何為物件添加值并洗掉當前值？

我有這樣的資料：

users = [{
    "emp_id": 1,
    "user": {
        "emp_full_name": "Test",
        "emp_email": "[email protected]",
        "emp_phone_no": null,
        "preferred_work_type": null
    },
    "hashtag": {
        "id": 1,
        "name": "NodeJs",
        "hashtag_group_id": 1
    },
    "difficulty": "HARD"
}, {
    "emp_id": 2,
    "user": {
        "emp_full_name": "test2",
        "emp_email": "[email protected]",
        "emp_phone_no": null,
        "preferred_work_type": null
    },
    "hashtag": {
        "id": 1,
        "name": "NodeJs",
        "hashtag_group_id": 1
    },
    "difficulty": "EASY"
}, {
    "emp_id": 1,
    "user": {
        "emp_full_name": "Test",
        "emp_email": "[email protected]",
        "emp_phone_no": null,
        "preferred_work_type": null
    },
    "hashtag": {
        "id": 4,
        "name": "Javascript",
        "hashtag_group_id": 1
    },
    "difficulty": "HARD"
}]

我想將hashtag具有相同emp_id. 如果emp_id有多個資料，則應洗掉emp_id具有單個資料的資料。hashtag

所以基本上這就是我所期望的：

[{
    "emp_id": 1,
    "user": {
        "emp_full_name": "Test",
        "emp_email": "[email protected]",
        "emp_phone_no": null,
        "preferred_work_type": null
    },
    "hashtag": [{
        "id": 1,
        "name": "NodeJs",
        "hashtag_group_id": 1
    }, {
        "id": 4,
        "name": "Javascript",
        "hashtag_group_id": 1
    }],
    "difficulty": "HARD"
}, {
    "emp_id": 2,
    "user": {
        "emp_full_name": "test2",
        "emp_email": "[email protected]",
        "emp_phone_no": null,
        "preferred_work_type": null
    },
    "hashtag": {
        "id": 1,
        "name": "NodeJs",
        "hashtag_group_id": 1
    },
    "difficulty": "EASY"
}]

如何轉換這樣的資料？

我不知道如何解決這個問題，我嘗試使用filter(), 并map()使用一些驗證條件，但無法讓它作業。

uj5u.com熱心網友回復：

您可以.reduce()使用.findIndex(). 試試這個

let users = '[{"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "[email protected]", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "HARD"}, {"emp_id": 2, "user": {"emp_full_name": "test2", "emp_email": "[email protected]", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 1, "name": "NodeJs", "hashtag_group_id": 1 }, "difficulty": "EASY"}, {"emp_id": 1, "user": {"emp_full_name": "Test", "emp_email": "[email protected]", "emp_phone_no": null, "preferred_work_type": null }, "hashtag": {"id": 4, "name": "Javascript", "hashtag_group_id": 1 }, "difficulty": "HARD"} ]';
users = JSON.parse(users)

users = users.reduce((arr, o) => {
    let idx = arr.findIndex(({emp_id}) => emp_id === o.emp_id);
    if( idx !== -1 ) arr[idx].hashtag = [].concat(arr[idx].hashtag, o.hashtag)
    else arr.push(o)

    return arr;
}, []);

console.log(users)

uj5u.com熱心網友回復：

您可以創建一個由該鍵鍵入的 Mapemp_id并通過該鍵收集用戶。當已有條目時，擴展hashtagusing [].concat。如果它還不是一個陣列，這將創建一個陣列。

const users = [{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "[email protected]","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "HARD"},{"emp_id": 2,"user": {"emp_full_name": "test2","emp_email": "[email protected]","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 1,"name": "NodeJs","hashtag_group_id": 1},"difficulty": "EASY"},{"emp_id": 1,"user": {"emp_full_name": "Test","emp_email": "[email protected]","emp_phone_no": null,"preferred_work_type": null},"hashtag": {"id": 4,"name": "Javascript","hashtag_group_id": 1},"difficulty": "HARD"}];

const map = new Map;
for (const user of users) {
    const match = map.get(user.emp_id);
    if (match) match.hashtag = [].concat(match.hashtag, user.hashtag);
    else map.set(user.emp_id, {...user});
}
const result = [...map.values()];

console.log(result);

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