ES6 换种思路处理数据

摘要：看完本文，希望可以写出更加漂亮简洁函数式的代码可以用来汇总数据的初始值把一个对象数组变成一个以数组中各个对象的为属性名，对象本身为属性值的对象。

Handle javascript data structures with map/reduce

看完本文，希望可以写出更加漂亮、简洁、函数式的代码?

reduce 可以用来汇总数据

const customer = [
  {id: 1, count: 2},
  {id: 2, count: 89},
  {id: 3, count: 1}
];
const totalCount = customer.reduce((total, item) =>
  total + item.count,
  0 // total 的初始值
);
// now totalCount = 92

把一个对象数组变成一个以数组中各个对象的 id 为属性名，对象本身为属性值的对象。haoduoshipin

let products = [
  {
    id: "123",
    name: "苹果"
  },
  {
    id: "345",
    name: "橘子"
  }
];

const productsById = products.reduce(
  (obj, product) => {
    obj[product.id] = product
    return obj
  },
  {}
);

console.log("result", productsById);

map 可以理解为是数组的转换器，依次对数组中的每个元素做变换进而得到一个新的数组。

const integers = [1, 2, 3, 4, 6, 7];
const twoXIntegers = integers.map(i => i*2);
// twoXIntegers are now [2, 4, 6, 8, 12, 14]
// integers数组并不会受到影响

筛选出数组中的个别元素

const posts = [
  {id: 1, title: "Title 1"},
  {id: 2, title: "Title 2"},
];
// find the title of post whose id is 1
const title = posts.find(p => p.id === 1).title;

唉~ 使用了半年的 es6才发现有这么好用的东西，译者傻缺还像下面这么写过呢

const posts = [
  {id: 1, title: "Title 1"},
  {id: 2, title: "Title 2"},
];

const title = posts.filter(item => item.id === 1)[0].title;

筛选出数组中某些符合条件的元素组成新的数组

const integers = [1, 2, 3, 4, 6, 7];
const evenIntegers = integers.filter(i => i % 2 === 0);
// evenIntegers are [2, 4, 6]

请大家自行思考下filter和find的区别

const arr1 = [1, 2, 3, 4, 5];
const arr2 = [6, 7, 8, 9, 0];
const arrTarget = [...arr1, ...arr2];
// [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

function operation(query, option = {}) {
    const param = {...query, ...option};
    // ....
    return param;
}
let opt = {startTime: 123455555, endTime: 113345555};
let q = {name: "一步", age: "xxx"};
operation(q, opt);
// {name: "一步", age: "xxx", startTime: 123455555, endTime: 113345555}

对象是引用传参的，所以函数内部应该尽可能的保证传入的参数不受到污染。

const dynamicKey = "wearsSpectacles";
const user = {name: "Shivek Khurana"};
const updatedUser = {...user, [dynamicKey]: true};
// updatedUser is {name: "Shivek Khurana", wearsSpectacles: true}

const params = {color: "red", minPrice: 8000, maxPrice: 10000};
const query = "?" + Object.keys(params)
  .map(k =>
    encodeURIComponent(k) + "=" + encodeURIComponent(params[k])
  )
  .join("&")
;
// encodeURIComponent encodes special characters like spaces, hashes
// query is now "color=red&minPrice=8000&maxPrice=10000"

const posts = [
  {id: 13, title: "Title 221"},
  {id: 5, title: "Title 102"},
  {id: 131, title: "Title 18"},
  {id: 55, title: "Title 234"}
];
// to find index of element with id 131
const requiredIndex = posts.map(p => p.id).indexOf(131);

更加优雅的写法呱呱呱提供：

const posts = [
  {id: 13, title: "Title 221"},
  {id: 5, title: "Title 102"},
  {id: 131, title: "Title 18"},
  {id: 55, title: "Title 234"}
];
const index = posts.findIndex(p => p.id === 131)

const user = { name: "Shivek Khurana", age: 23, password: "SantaCl@use" };
const userWithoutPassword = Object.keys(user)
    .filter(key => key !== "password")
    .map(key => ({[key]: user[key]}))
    .reduce((accumulator, current) => ({ ...accumulator, ...current }), {});

这里我认为原作者有点为了函数式编程而函数式了，下面是我的解决方案：

const user = {name: "Shivek Khurana", age: 23, password: "SantaCl@use"};
const newUser = {...user};
delete newUser.password;
// {name: "Shivek Khurana", age: 23}

更现代的写法YiHzo提供: ?????

const user = {name: "Shivek Khurana", age: 23, password: "SantaCl@use"};
// 利用对象的解构，取出非password的所有字段
const {password, ...newUser} = user

以上代码片段的共同原则：不改变原数据。希望大家的代码都可以尽可能的简洁，可维护?。

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