javascript一些小但实用的函数

获取本年度第一天

            nowYear() {
                const sjc = Date.now()
                const date = new Date(parseInt(sjc)).toLocaleString()
                let D = parseInt(date.split('/')[2]) + ''
                let M = date.split('/')[1]
                const Y = date.split('/')[0]
                return ('' + Y + '-01-01')
            },

获取当天时间

nowDate() {
                const timeStamp = Date.now()
                const date = new Date(parseInt(timeStamp)).toLocaleString()
                let D = parseInt(date.split('/')[2]) + ''
                let M = date.split('/')[1]
                const Y = date.split('/')[0]
                if (M.length == 1) {
                    M = '0' + M
                }
                if (D.length == 1) {
                    D = '0' + D
                }
                return ('' + Y + '-' + M + '-' + D)
            },
           

获取昨日时间

            
            afterDate() {
                const timeStamp = Date.now()
                const date = new Date(parseInt(timeStamp) - 86400000).toLocaleString()
                let D = parseInt(date.split('/')[2]) + ''
                let M = date.split('/')[1]
                const Y = date.split('/')[0]
                if (M.length == 1) {
                    M = '0' + M
                }
                if (D.length == 1) {
                    D = '0' + D
                }
                return ('' + Y + '-' + M + '-' + D)
            },

一行代码将a,b两个变量交换值

let a = 10
let b = 20
b = [a,a=b][0]
// a = 20
// b = 10

最短的代码实现数组去重

[...new Set([1, '1', 2, 1, 1, 'a', 'b'])]

'&&'的的隐藏用法

{ variable && function（fnc）{} } 
相当于↓
if(variable){
function（fnc）{}
}
// '&&'相当于一扇门，只有值是true才会打开，并执行接下来的代码

在控制台输入以下代码，会显得你很nb

console.log(([][[]]+[])[+!![]]+([]+{})[!+[]+!![]])