1.forEach(),遍歷數組的每個元素
let arrFor = ['muzi','digbig','muzidigbig','lucky'] //forEach(),遍歷數組的每個元素 arrFor.forEach((item,index) => { console.log(`${index}---${item}`) })2.map(參數為回調函數)函數，遍歷數組每個元素，并回調操作，需要返回值，返回值組成新數組，原數組不變；
let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}] const arrMap = arr.map((item,index) => { return { id:item.id, name:item.name, sex:'男' } }) console.log(arrMap)
3.filter(參數為回調函數)函數：過濾通過條件的元素組成一個新數組，原數組不變；
let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}] const arrFilter = arr.filter((item,index) => { return item.id >= 2; }) console.log(arrFilter)
4.some(參數為回調函數)函數，遍歷數組中是否有符合條件的函數，返回布爾值；
let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}] const arrSome = arr.some((item,index) => { return item.id === 5 }) console.log(arrSome)5.every(參數為回調函數)函數，遍歷數組是否每個元素都符合條件，返回布爾值；
let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}] const arrEvery = arr.every((item,index) => { return item.id >= 1 }) console.log(arrEvery)6.find()函數，數組中的每個元素都執行這個回調函數；返回第一個滿足條件的元素 之后的元素就不在調用；沒有符合的返回undefined；并沒有改變數組的原始值。
let arr = [{id:1,name:'muzi'},{id:2,name:'digbig'},{id:3,name:'muzidigbig'},{id:4,name:'lucky'}] const arrFind = arr.find((item,index) => { return item.id === 1 }) console.log(arrFind)7.reduce(),合并二維數組
var twoArr = [['mu','zi'],['dig','big'],['lucky','jiji']]; var oneArr = twoArr.reduce(function(total,currentValue){ // console.log(total) return total.concat(currentValue); },[]) console.log(oneArr);//["mu", "zi", "dig", "big", "lucky", "jiji"]總結
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