刪除對象數組中的對象鍵和值

Question

我的代碼 =

這是我的示例數據。

const songs = [
  {
    "title": "Just Once",
    "asset_id": "1f7e0fd8-db21-4c28-b9e1-eb0295af198c",
    "sort": 1,
    "performers": [
      {
        "last_name": "John",
        "first_name": "Doe",
        "group": {
          "group_id": "1e5f73fa-ffe8-4c70-a83b-84e7bf985b25",
          "dept_short_name": "PAO",
          "dept_long_name": "Public Affairs Office"
        },
        "email": "john@doe.com"
      }
    ]
  }
]

    const newTest= songs.map(( {...song} ) => (
        song.performers.map(({...group}) => group_id = group.group_id)
    ))

結果我得到了這個：

我應該只從組對象中刪除dept_short_name和dept_long_name ，而group_id將保留，歌曲的其余數據應保持不變。

結果應該是這樣：

[
  {
    "title": "Just Once",
    "asset_id": "1f7e0fd8-db21-4c28-b9e1-eb0295af198c",
    "sort": 1,
    "performers": [
      {
        "last_name": "John",
        "first_name": "Doe",
        "group_id":"1e5f73fa-ffe8-4c70-a83b-84e7bf985b25"
        "email": "john@doe.com"
      }
    ]
  }
]

Answer 1

您需要在兩個地圖中重新創建一個對象。

 const songs = [{"title":"Just Once","asset_id":"1f7e0fd8-db21-4c28-b9e1-eb0295af198c","sort":1,"performers":[{"last_name":"John","first_name":"Doe","group":{"group_id":"1e5f73fa-ffe8-4c70-a83b-84e7bf985b25","dept_short_name":"PAO","dept_long_name":"Public Affairs Office"},"email":"john@doe.com"}]}] const updated = songs.map(({ performers, ...song }) => ({ ...song, // recreate song object performers: performers.map(({ group: { group_id }, ...performer }) => ({ ...performer, // recreate performer object group_id })) })) console.log(updated)

Answer 2

  {
    "title": "Just Once",
    "asset_id": "1f7e0fd8-db21-4c28-b9e1-eb0295af198c",
    "sort": 1,
    "performers": [
      {
        "last_name": "John",
        "first_name": "Doe",
        "group": {
          "group_id": "1e5f73fa-ffe8-4c70-a83b-84e7bf985b25",
          "dept_short_name": "PAO",
          "dept_long_name": "Public Affairs Office"
        },
        "email": "john@doe.com"
      }
    ]
  }
]

window.addEventListener('load', function() {
 delete songs[0].performers[0].group.dept_short_name;
 delete songs[0].performers[0].group.dept_long_name;
 console.log(JSON.stringify(songs))
});

這可能會有所幫助。

Answer 3

const songs = [{
  title: "Just Once",
  asset_id: "1f7e0fd8-db21-4c28-b9e1-eb0295af198c",
  sort: 1,
  performers: [{
    last_name: "John",
    first_name: "Doe",
    group: {
      group_id: "1e5f73fa-ffe8-4c70-a83b-84e7bf985b25",
      dept_short_name: "PAO",
      dept_long_name: "Public Affairs Office"
    },
    email: "john@doe.com"
  }]
}];
songs.map(({ performers }, ind) => {
 performers.map(({ group }, ind) => {
  performers[ind].group_id = group.group_id;
 });

 delete performers[ind].group;
});

console.log(songs);

Answer 4

使用delete關鍵字：delete performerce ["keyname"] 返回布爾值