从对象数组中筛选符合条件的值

遍历当前数组中的键拿到另一个数组中包含相同键的当前数组对象

const arr = [1,2,3,4,5,6,7]
const list = [
	{
    
    openId: 1, timelineId: 1, showNo: 1, uid: 1},
	{
    
    openId: 2, timelineId: 1, showNo: 1, uid: 1},
	{
    
    openId: 9, timelineId: 1, showNo: 1, uid: 1},
	{
    
    openId: 4, timelineId: 1, showNo: 1, uid: 1},
	{
    
    openId: 5, timelineId: 1, showNo: 1, uid: 1}
]
const params = list.filter(item=> arr.indexOf(item.openId) > -1)
console.log(params)

将两个对象数组根据相同的索引index合并为一个数组

this.currentTotalList = this.totalList.map((item, index) => ({
    
     ...item, ...daysList[index] }))

将两个对象数组根据相同的键值合并为一个数组

let currentEveryList = this.everyList.map(item => ({
    
    ...item, ...signList.filter(s => s.signDate === item.signDate)[0]}))

从当前数组中筛选符合条件的值

this.materialss = this.materials.filter(item => item.categoryId === this.curTab.categoryId)