Object.keys,Object.values(),Obje

2020-03-17  本文已影响0人  小棋子js

**Object.keys()** (取对象名)方法会返回一个由一个给定对象的自身可枚举属性组成的数组,数组中属性名的排列顺序和使用[for...in]循环遍历该对象时返回的顺序一致 。
如果你想获取一个对象的所有属性,,甚至包括不可枚举的,请查看[Object.getOwnPropertyNames]

在实际开发中,我们有时需要知道对象的所有属性;
ES5 引入了Object.keys方法,成员是参数对象自身的(不含继承的)所有可遍历( enumerable )属性的键名。

var data={a:1,b:2,c:9,d:4,e:5};
    console.log(Object.keys(data));//["a", "b", "c", "d", "e"]
    Object.keys(data).map((key,item)=>{
        console.log(key,data[key]);//key=>属性名    data[key]=>属性值
});
var str = 'ab1234';
console.log(Object.keys(obj));  //[0,1,2,3,4,5]
var arr = ["a", "b", "c"];
    console.log(Object.keys(arr)); // console: ["0", "1", "2"]
function Pasta(name, age, gender) {
      this.name = name;
      this.age = age;
      this.gender = gender;
      this.toString = function () {
            return (this.name + ", " + this.age + ", " + this.gender);
    }
}

console.log(Object.keys(Pasta)); //console: []

var spaghetti = new Pasta("Tom", 20, "male");
console.log(Object.keys(spaghetti)); //console: ["name", "age", "gender", "toString"]

扩展

  1. Object.values()取对象值
    Object.values方法返回一个数组,成员是参数对象自身的(不含继承的)所有可遍历( enumerable )属性的键值。Object.values()和Object.keys()是相反的操作,把一个对象的值转换为数组
var data={a:1,b:2,c:9,d:4,e:5};
    console.log(Object.values(data));//[1, 2, 9, 4, 5]
    Object.values(data).map((key,item)=>{
        console.log(key,data[key]);//key=>属性值    data[key]=>undefined
});
  1. Object.entries()把对象转换为一个键值对列表
    Object.entries方法返回一个数组,成员是参数对象自身的(不含继承的)所有可遍历( enumerable )属性的键值对数组
var data={a:1,b:2,c:9,d:4,e:5};
    console.log(Object.entries(data));//[["a", 1], ["b", 2],["c", 9],["d", 4], ["e", 5]]
    Object.entries(data).map((key,item)=>{
        console.log(key);// key=>["a", 1],["b", 2],["c", 9],["d", 4],["e", 5]   
});
// 转换
console.log(Object.entries({ name: 'JAY', age: 41 })); 
//[ [ 'name', 'JAY' ], [ 'age', 41 ] ]

console.log(Object.entries([1,2,3,4])) 
//[ [ '0', 1 ], [ '1', 2 ], [ '2', 3 ], [ '3', 4 ] ]

console.log(Object.entries('banana'))  
//[ [ '0', 'b' ],[ '1', 'a' ],[ '2', 'n' ],[ '3', 'a' ],[ '4', 'n' ],[ '5', 'a' ] ]

console.log(Object.entries(1)) //传入数字和浮点数返回 []

// 将OBject 转换为map
const obj = { name: 'JAY', age: 41 };
const map = new Map(Object.entries(obj))
console.log(map) 
// Map { 'name' => 'JAY', 'age' => 41 }
  1. Object.fromEntries() 方法把键值对列表转换为一个对象。
    Object.fromEntries 列子
// 转化map
const map = new Map([
  ['name','周杰伦'],['age',36],
])

console.log(Object.fromEntries(map)) 
// {name: "周杰伦", age: 36}

// 转化array
const arr = [
  ['0', 'a'],
  ['1', 'b'],
  ['2', 'c']
];
console.log(Object.fromEntries(arr)) 
// {0: "a", 1: "b", 2: "c"}

// 将对象中的属性数值统一*n
const obj1 = {a:1,b:2,c:3,d:'isD'}
const obj2 = Object.fromEntries(
  Object.entries(obj1).map(([key,val])=>{
    if(typeof val === 'number')return [key,val*2]
    else return [key,val]
  })
)
console.log(obj2) 
//{a: 2, b: 4, c: 6, d: "isD"}

// format url 

const str = "age=zhoujielun&&name=16"
const ar = new URLSearchParams(str)
console.log(Object.fromEntries(ar)) 
//{age: "zhoujielun", name: "16"}

简易实现

Object.entries
const entries = (inArg)=>{
  if(Array.isArray(inArg)){
    return inArg.map((x,index)=>[`${index}`,x])
  }
  if(Object.prototype.toString.call(inArg) === `[object Object]`){
   return Object.keys(inArg).map(y=>[y,inArg[y]])
  }
  if(typeof inArg === 'number')return []
  throw 'Cannot convert argument to object'
} 

// test
console.log(entries(1)) 
// []

console.log(entries([1,2,3])) 
// [ [ '0', 1 ], [ '1', 2 ], [ '2', 3 ] ]

console.log(entries({age:12,name:'ysss'})) 
// [ [ 'age', 12 ], [ 'name', 'ysss' ] ]

if(!Object.entries)Object.entries = entries;
Object.fromEntries

const fromEntries = (arrArg)=>{
  // Map
  if (Object.prototype.toString.call(arrArg) === '[object Map]') {
    const resMap = {};
    for (const key of arrArg.keys()) {
      resMap[key] = arrArg.get(key);
    }
    return resMap;
  }
  // array
  if(Array.isArray(arrArg)){
    const resArr = {}
    arrArg.map(([key,value])=>{
      resArr[key] =  value
    })
    return resArr
  }
  throw 'Uncaught TypeError: argument is not iterable';
}

// test 
const map = new Map([
  ['name', '周杰伦'],
  ['age', 36]
]);
console.log(fromEntries(map));
 // {name: "周杰伦", age: 36}

const arr = [
  ['0', 'a'],
  ['1', 'b'],
  ['2', 'c']
];
console.log(fromEntries(arr)) 
// {0: "a", 1: "b", 2: "c"}

// polyfill
if(!Object.fromEntries)Object.fromEntries = fromEntries 

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