es6（二）：解构赋值

1 let [a,b,c]=[1,10,100]
2     console.log(a,b,c)//1 10 100
3     //等式两边"模式"进行匹配，从而进行赋值
4     let [i,[[[j]],k]]=[1,[[[2]],3]]//这种看看就好，知道这样也可以执行就行
5     console.log(i,j,k)//1 2 3
6     
7     let [,,x]=[1,2,3]
8     console.log(x)//3

1 let [i1,i2]=[]
2     console.log(i1,i2)//undefined undefined
3     let [i3,i4]=[10]
4     console.log(i3,i4)//10 undefined

1 let [i5,i6]=[1,2,3]
2     console.log(i5,i6)//1 2

1 let [i7,i8]=1 //1 is not iterable
2 let [i9] =true//true is not iterable

1 let [i10,i11='hi']=[100]
2     console.log(i10,i11)//100 'hi'

1 // let [i111=i12,i12=10]=[]//报错，i111=i12 此时i12还未申明
2     // 而：
3     let [i13=i14,i14]=[1,2]
4     console.log(i13,i14)

 1 let {foo,bar}={foo:'name',bar:'age'}
 2     console.log(foo,bar)
 3     //但其实上面解构赋值是下面的缩写
 4     let {foo1:foo1,bar1:bar1}={foo1:'name',bar1:'age'}
 5     console.log(foo1,bar1)
 6     // 如果继续变换
 7     let {foo2:name,bar2:age}={foo2:'name',bar2:'age'}
 8     // console.log(foo2,bar2)//报错
 9     console.log(name,age)//正确写法
10     // 即foo2是模式，name才是赋值的变量

1 let a10=10,b100=100;
2     [a10,b100]=[b100,a10]
3     console.log(a10,b100)//100 10

 1 function test10(){
 2       return [1,2,3]
 3     }
 4     function test100(){
 5       return {
 6         nameT:'apple',
 7         ageT:100
 8       }
 9     }
10     let [j1,j2,j3]=test10()
11     console.log(j1,j2,j3)//1 2 3 
12     let {nameT,ageT}=test100()
13     console.log(nameT,ageT) //'apple' 100

1 function getJSON(){
2       return {
3         "app1":100,
4         "blue":"hello blue",
5         "data": 110
6       }
7     }
8     let {app1,blue,data:number}=getJSON()
9     console.log(app1,blue,number)//100 "hello blue" 110