Javascript Generator

n日後の営業日を返す. 長い...

code: dayafter.js

// day and holidays format yyyy-mm-dd

function isBusinessDay(day, holidays) {

if (holidays.includes(day)) return false

return true

}

// day format yyyy-mm-dd

function getTomorrow(day) {

const today = new Date(day)

const tomorrow = new Date(today)

tomorrow.setDate(tomorrow.getDate() + 1)

return formatDate(tomorrow)

}

// 日付をYYYY-MM-DDの書式で返すメソッド

function formatDate(dt) {

var y = dt.getFullYear();

var m = ('00' + (dt.getMonth() + 1)).slice(-2);

var d = ('00' + dt.getDate()).slice(-2);

return (y + '-' + m + '-' + d);

}

// generator

function* gen(holidays, day, later) {

let count = 0

const exceptionalNum = 100

for (let i = 1; i < exceptionalNum; i++) {

day = getTomorrow(day)

//console.log(day)

//console.log('holiday')

;//do nothing

} else {

count++

}

if (count == later) break;

if( i > exceptionalNum) throw 'gen() does not work'

yield count

}

return day

}

async function main() {

const today = await fetch('now.json').then(r => r.json())

const today_str = today.today

const holidays = await fetch('holidays.json').then(r => r.json())

const g = gen(holidays, "2021-12-29", 5)

let result = g.next();

while (!result.done) {

result = g.next()

}

document.getElementById('calendar').setAttribute("min", result.value)

console.log(result)

}

document.addEventListener('DOMContentLoaded', function () {

main().then(r => console.log('done'))

})