时间:2021-07-01 10:21:17 帮助过:3人阅读
1.forEach(),遍历数组的每个元素
let arrFor = ['muzi','digbig','muzidigbig','lucky'] //forEach(),遍历数组的每个元素 arrFor.forEach((item,index) => { console.log(`${index}---${item}`) })
2.map(参数为回调函数)函数,遍历数组每个元素,并回调操作,需要返回值,返回值组成新数组,原数组不变;
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(参数为回调函数)函数:过滤通过条件的元素组成一个新数组,原数组不变;
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(参数为回调函数)函数,遍历数组中是否有符合条件的函数,返回布尔值;
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(参数为回调函数)函数,遍历数组是否每个元素都符合条件,返回布尔值;
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()函数,数组中的每个元素都执行这个回调函数;返回第一个满足条件的元素 之后的元素就不在调用;没有符合的返回undefined;并没有改变数组的原始值。
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(),合并二维数组
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"]
总结
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