Question

我正在VueJS做一个项目，我有一个数据数组，其中包含许多对象。

这些对象是从PHP后端提取的，由

等值组成

id：2123 名称：＆＃34;名称值＆＃34; 状态：＆＃34;活跃＆＃34; account_id：＆＃34; 2KGGALS2353255＆＃34;

想象一下，我想通过键名将它们拆分成一个类似的数组，但我希望有一个由两个子对象组成的父对象

[
    0: {
     core: {
       id: 2123
       name: "Name Value"
     },
     extra: {
       status: "active",
       account_id: "2KGGALS2353255" 
     }
]

问题是如何使用Javascript实现这一目标？我不想事先修改PHP中的数据，除非这是Javascript中非常缺乏的。我可以使用VueJS和Lodash。

Answer 1

我一直在寻找lodash的pick()方法。

https://lodash.com/docs/4.17.4#pick

Answer 2

这应该适用于您的目的

function separate(obj, keys) {
  let target = {}, rest = {};
  Object.keys(obj).forEach(function(key) {
    if (keys.includes(key)) {
      target[key] = obj[key];
    } else {
      rest[key] = obj[key];
    }
  });
  return { target: target, rest: rest };
}

let stuff = { 
  id: 2123,
  name: "Name Value",
  status: "active",
  account_id: "2KGGALS2353255" 
};

let separated = separate(stuff, ['id', 'name']);

console.log({
  core: separated.target,
  extra: separated.rest
});

Answer 3

使用ES6的object destructuring和object rest spread提案（需要babel transform），您可以Array#map将数组放入所需格式的新对象数组中：< / p>

const arr = [{"id":1,"name":"Name1","status":"active","account_id":"2KGGALS2353255"},{"id":2,"name":"Name2","status":"active","account_id":"4ABCLS2353255"},{"id":3,"name":"Name3","status":"active","account_id":"6LMNALS2353255"}];

const result = arr.map(({ id, name, ...extra }) => ({
  core: {
    id,
    name
  },
  extra
}));

console.log(result);

你可以使用lodash的_.pick()来获取核心，_.omit()来获得额外的：

var arr = [{"id":1,"name":"Name1","status":"active","account_id":"2KGGALS2353255"},{"id":2,"name":"Name2","status":"active","account_id":"4ABCLS2353255"},{"id":3,"name":"Name3","status":"active","account_id":"6LMNALS2353255"}];

var result = arr.map(function(obj) {
  return {
    core: _.pick(obj, ['id', 'name']),
    extra: _.omit(obj, ['id', 'name'])
  };
});

console.log(result);

<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.4/lodash.min.js"></script>

通过名称将对象的属性拉入另一个对象

3 个答案: