Question

我有以下对象：

{
  "_id": "583c4e054c99d310f543b11e",
  "cellphone": "123456",
  "password": "$2a$10$d3SeD5CCzTo6wBR/4SGUu.i7vUvX98N1NlpBTwDdWCRrIYcVwWmCO",
  "fullname": "some name",
  "gender": "male",
  "avatar": {
    "_id": "583c4e054c99d310f543b11d",
    "__v": 0,
    "url": "",
    "likesCount": 0,
    "description": "Hey guys"
  },
  "__v": 0
}

通过下一个代码，我想创建一个没有mongodb（__v）属性的新对象，并将_id重命名为id。

  let plainUser = {
    id: user._id,
    ...user.toObject(),
  }

  delete plainUser._id;
  delete plainUser.password;
  delete plainUser.__v;
  delete plainUser.avatar._id;
  delete plainUser.avatar.__v;

  return res.send(plainUser);

我想我能以不同的方式做到这一点。任何建议如何使用ES6正确改进我的代码？

Answer 1

最好明确保留哪些属性，而不是指定应删除哪些属性。 delete效率很低。所以只需去

return res.send({
    id: user._id,
    cellphone: user.cellphone,
    fullname: user.fullname,
    gender: user.gender,
    avatar: {
        url: user.avatar.url,
        likesCount: user.avatar.likesCount,
        description: user.avatar.description
    }
});

为了简化，请查看One-liner to take some properties from object in ES 6的方法：

const {_id: id, cellphone, fullname, gender} = user;
const avatar = (({url, likesCount, description}) => ({url, likesCount, description}))(user.avatar);
return res.send({id, cellphone, fullname, gender, avatar});

Answer 2

这是一个规范化对象属性的函数：

function normalizeObjectKeys(obj) {
    let normilizedObj = {};
    Object.keys(obj).map(key => {
        let newKey = key.replace(/_+/, '');
        normilizedObj[newKey] = ('object' !== typeof obj[key]) ? obj[key] : normalizeObjectKeys(obj[key]);
    });
    return normilizedObj;
}

示例：

normalizeObjectKeys({
  "_id": "583c4e054c99d310f543b11e",
  "cellphone": "123456",
  "password": "$2a$10$d3SeD5CCzTo6wBR/4SGUu.i7vUvX98N1NlpBTwDdWCRrIYcVwWmCO",
  "fullname": "some name",
  "gender": "male",
  "avatar": {
    "_id": "583c4e054c99d310f543b11d",
    "__v": 0,
    "url": "https://pp.vk.me/c631525/v631525614/2b398/kbI5QohgEgQ.jpg",
    "likesCount": 0,
    "description": "Hey guys"
  },
  "__v": 0
});

如何正确分配对象属性是JS

2 个答案: