ES6解构赋值

ES6结构赋值

　　左右两边结构一样，声明和赋值不能分开必须要再同一行的代码里面

　　1、数组模型的解构

let arr = [1, 2, 3];
let a = arr[0];
let b = arr[1];
let c = arr[2];
console.log(a, b, c); // 1 ，2 ，3

let [a, b, c] = [1, 2, 3];
console.log(a, b, c); // 1 2 3

// 嵌套
let [a, [[b], c]] = [1, [[2], 3]];
console.log(a, b, c); // 1 2 3

// 忽略
let [a, , b] = [1, 2, 3];
console.log(a, b); // 1  3

// 不完全解构
let [a = 1, b] = []; 
console.log(a, b); // a = 1, b = undefined

// 剩余运算符
let {a, b, ...rest} = {a: 10, b: 20, c: 30, d: 40};
// a = 10
// b = 20
// rest = {c: 30, d: 40}

　　2、json解构

let {a, b, c} = {a: 1, b: 2, c: 3};
console.log(a, b, c); // 1 2 3

let [{a, b}, [n1, n2], num, str] = [{a: 1, b: 2}, [3, 4], 5, '6'];
console.log(a, b, n1, n2, num, str); // 1 2 3 4 5 "6"

let [json, arr, num, str] = [{a: 1, b: 2}, [3, 4], 5, '6'];
console.log(json, arr, num, str); // {a: 1, b: 2} (2) [3, 4] 5 "6"

　　3、解构，默认值

let {a = 10, b = 5} = {a: 3};
// a = 3; b = 5;
let {a: aa = 10, b: bb = 5} = {a: 3};
// aa = 3; bb = 5;

　　当解构模式有匹配结果，且匹配结果是 undefined 时，会触发默认值作为返回结果。

let [a = 3, b = a] = [];     // a = 3, b = 3 a 与 b 匹配结果为 undefined ，触发默认值：a = 3; b = a =3
let [a = 3, b = a] = [1];    // a = 1, b = 1 a 正常解构赋值，匹配结果：a = 1，b 匹配结果 undefined ，触发默认值：b = a =1
let [a = 3, b = a] = [1, 2]; // a = 1, b = 2 a 与 b 正常解构赋值，匹配结果：a = 1，b = 2

　　4、注意事项

// 左边为数组，右边为json，两边的结构不一样
let [a, b] = {a: 12, b: 34};
console.log(a, b); // 出错

// 右边类型错误，{12， 34} json写法错误
let {a, b} = {12, 34};
console.log(a, b); // Uncaught SyntaxError: Unexpected token 

// 声明和赋值分开
let [a, b];
[a , b] = {12, 34};
console.log(a, b); // Uncaught SyntaxError: Missing initializer in destructuring declaration