**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。
- 传入对象，返回属性名
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
});
- 传入字符串，返回索引
var str = 'ab1234';
console.log(Object.keys(obj)); //[0,1,2,3,4,5]
- 传入数组 返回索引
var arr = ["a", "b", "c"];
console.log(Object.keys(arr)); // console: ["0", "1", "2"]
- 构造函数 返回空数组或者属性名
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
- 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
});
- 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 }
- 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
