这是我第一次用 Javascript 编写代码，我在其他编码语言中也没有更好，但我设法让代码工作（有点）。
我认为这段代码可以更短，但还有另一个我不太明白的问题。 如果有人能指出我的错误并告诉我更简单的方法，这对未来真的很有帮助。
var main = function() {
//put the day of payment in here
const payday = 25;
//This code is needed for the following Calculation
const now = new Date();
currentDay = now.getDate();
const nextMonth = new Date(now.getFullYear(), now.getMonth() + 1, 1);
const diffDays = Math.ceil(Math.abs(nextMonth.getTime() - now.getTime()) / (1000 * 60 * 60 * 24));
now.setMonth(now.getMonth() + 1);
now.setDate(0);
const daysInCurrentMonth = now.getDate();
//Not necessary
//console.log('Days in current month: ' + daysInCurrentMonth);
//console.log('Days until next month: ' + diffDays);
//Calculation of days left until payment
//Here 25th is the day of Payment you can change that to your liking
const daysTilPayment = Math.ceil(diffDays - (daysInCurrentMonth - payday) - 1 );
//Folowing code is to determine if the payday is on a weekend
//Also it changes the paydate to 23th since thats how banks operate Here in Switzerland
//If your paydate will be changed to the closest workday (Friday) Then
//change the 2 on Rules for Saturdays to 1
//If that doesnt apply to you at all remove the 'Rules for Saturdays' + 'Rules for Sundays'
//Sunday = 0 and Saturday = 6
//to determine weekday of payment in current month
const d = new Date();
d.setDate(d.getDate() + daysTilPayment);
let paymentDay = d.getDay();
//to determine weekday of payment in next month
const thenextMonth = new Date (d.setMonth(d.getMonth() +4));
let paymentDay2 = d.getDay();
{
if (paymentDay >5)
//Rules for Saturdays
//To subtract 2
output = Math.ceil(daysTilPayment - 2);
//End of rules for Saturdays
else if (paymentDay <1)
//Rules for Sundays
//To subtract 2
output = Math.ceil(daysTilPayment - 2);
//End of rules for Sundays
else
//Rules for normal days
//does nothin
output = daysTilPayment
//End of rules for normal days
}
//this made the repetition above necessary
//it is to get rid of the countdown going negativ after payday
if (output < 0)
//output = Math.ceil(daysInCurrentMonth - currentDay + payday-2);
{
if (paymentDay2 >5)
//Rules for Saturdays
{//To put a 0 in front of single digit numbers and subtract 2
if (daysTilPayment+ daysInCurrentMonth - currentDay + payday -2 < 10)
output = ': 0'+Math.ceil(daysInCurrentMonth - currentDay + payday-2);
else
output = ': '+Math.ceil(daysInCurrentMonth - currentDay + payday-2);
}
//End of rules for Saturdays
else if (paymentDay2 <1)
//Rules for Sundays
{//To put a 0 in front of single digit numbers and subtract 2
if (daysTilPayment+ daysInCurrentMonth - currentDay + payday -2 < 10)
output = ': 0'+Math.ceil(daysInCurrentMonth - currentDay + payday-2);
else
output = ': '+Math.ceil(daysInCurrentMonth - currentDay + payday-2);
}
//End of rules for Sundays
else
//Rules for normal days
{//Just to put a 0 in front of single digit numbers
if (daysTilPayment+ daysInCurrentMonth - currentDay + payday < 10)
output = ': 0'+Math.ceil(daysInCurrentMonth - currentDay + payday);
else
output = ': '+Math.ceil(daysInCurrentMonth - currentDay + payday);
}
//End of rules for normal days
}
else
{
if (paymentDay >5)
//Rules for Saturdays
{//To put a 0 in front of single digit numbers and subtract 2
if (daysTilPayment -2 < 10)
output = ': 0'+Math.ceil(daysTilPayment - 2);
else
output = ': '+Math.ceil(daysTilPayment - 2);
}
//End of rules for Saturdays
else if (paymentDay <1)
//Rules for Sundays
{//To put a 0 in front of single digit numbers and subtract 2
if (daysTilPayment -2 < 10)
output = ': 0'+Math.ceil(daysTilPayment - 2);
else
output = ': '+Math.ceil(daysTilPayment - 2);
}
//End of rules for Sundays
else
//Rules for normal days
{//Just to put a 0 in front of single digit numbers
if (daysTilPayment < 10)
output = ': 0'+daysTilPayment;
else
output = ': '+daysTilPayment;
}
//End of rules for normal days
}
//for debug purpose
//console.log(output);
return output.toLocaleLowerCase('de-DE');
}
因为我想在小部件上使用它，所以我返回本地小写，但输出有时是一个数学结果，无法以这种方式返回。 为了解决这个问题，我在前面放了一个 : ，以便输出是文本而不是数学结果。 它解决了这个问题，但现在小部件总是输出 :nn 而不仅仅是 nn 。 这没什么大不了的，我现在只想知道如何正确地完成它。
如果我胡言乱语，我很抱歉。 我真的不知道自己在做什么。
OP代码比较复杂，难以阅读。据我所知，它应该计算距离发薪日的天数，发薪日应该是该月的 25 日，或者如果是星期六或星期日，则为上一个星期五。
OP 代码中的问题太多，无法逐一解决。请参阅 在 JavaScript 中获取 2 个日期之间的差异？ 以及某些问题的链接副本。这是基于上述规则的简单解决方案。
/* Return next pay day.
*
* If defaultPayDate falls on Sat or Sun, set payDate to prior Fri
* If payDate is after baseDate, get next month's pay date
*
* @param {number} defaultPayDate - day of month for pay day
* @param {Date} baseDate - date to calculate days to pay day from
* @returns {Date} - next pay day
*/
function getNextPayday(defaultPayDate = 25, baseDate = new Date()) {
// Get payday for the baseDate month
let payDate = new Date(baseDate.getFullYear(), baseDate.getMonth(), defaultPayDate);
// If on Sat (6) or Sun (0), shift to prior Fri
if (!payDate.getDay() % 6) {
// Subtract 2 from Sun, 1 from Sat
payDate.setDate(payDate.getDate() - (payDate.getDay() + 2 % 7));
}
// If baseDate is after payDate, get next month's pay day
if (payDate.getDate() < baseDate.getDate()) {
payDate = getNextPayday(defaultPayDate, new Date(baseDate.getFullYear(), baseDate.getMonth() + 1, 1));
}
return payDate;
}
/* Days to next payday is not inclusive. Requires getNextPayday
*
* @param {number} defaultPayDate - day of month for pay day
* @param {Date} baseDate - date to calculate days to pay day from
* @returns {number} - days to next pay day (integer)
*/
function getDaysToNextPay(defaultPayDate = 25, baseDate = new Date()) {
let payDate = getNextPayday(defaultPayDate, baseDate);
// Use UTC to get differnce in days as UTC days are always exactly 8.64e7 ms long
let daysToPayday = (Date.UTC(payDate.getFullYear(), payDate.getMonth(), payDate.getDate()) -
Date.UTC(baseDate.getFullYear(), baseDate.getMonth(), baseDate.getDate())) / 8.64e7;
return daysToPayday;
}
// Example
let d = new Date();
console.log('Today is : ' + d.toDateString());
console.log('Next payday is : ' + getNextPayday(25, d).toDateString());
console.log('Days to next payday: ' + getDaysToNextPay(25, d));