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js中Array對象的常用遍歷方法詳解

 更新時(shí)間:2019年01月17日 11:53:40   作者:muzidigbig  
今天小編就為大家分享一篇關(guān)于js中Array對象的常用遍歷方法詳解,小編覺得內(nèi)容挺不錯(cuò)的,現(xiàn)在分享給大家,具有很好的參考價(jià)值,需要的朋友一起跟隨小編來看看吧

1.forEach(),遍歷數(shù)組的每個(gè)元素

  let arrFor = ['muzi','digbig','muzidigbig','lucky']
  //forEach(),遍歷數(shù)組的每個(gè)元素
  arrFor.forEach((item,index) => {
    console.log(`${index}---${item}`)
  })

2.map(參數(shù)為回調(diào)函數(shù))函數(shù),遍歷數(shù)組每個(gè)元素,并回調(diào)操作,需要返回值,返回值組成新數(shù)組,原數(shù)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(shù)為回調(diào)函數(shù))函數(shù):過濾通過條件的元素組成一個(gè)新數(shù)組,原數(shù)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(shù)為回調(diào)函數(shù))函數(shù),遍歷數(shù)組中是否有符合條件的函數(shù),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(shù)為回調(diào)函數(shù))函數(shù),遍歷數(shù)組是否每個(gè)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(shù),數(shù)組中的每個(gè)元素都執(zhí)行這個(gè)回調(diào)函數(shù);返回第一個(gè)滿足條件的元素 之后的元素就不在調(diào)用;沒有符合的返回undefined;并沒有改變數(shù)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(shù)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"]

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