我如何在JavaScript中计算出两个Date()对象的差异,而只返回差异中的月份数?
任何帮助都是最好的:)
我如何在JavaScript中计算出两个Date()对象的差异,而只返回差异中的月份数?
任何帮助都是最好的:)
当前回答
getMonthDiff(d1, d2) {
var year1 = dt1.getFullYear();
var year2 = dt2.getFullYear();
var month1 = dt1.getMonth();
var month2 = dt2.getMonth();
var day1 = dt1.getDate();
var day2 = dt2.getDate();
var months = month2 - month1;
var years = year2 -year1
days = day2 - day1;
if (days < 0) {
months -= 1;
}
if (months < 0) {
months += 12;
}
return months + years*!2;
}
其他回答
#这是我写的一段很好的代码,用于获取天数和月份 从给定日期开始
把你的手
/** * Date a end day * Date b start day * @param DateA Date @param DateB Date * @returns Date difference */ function getDateDifference(dateA, DateB, type = 'month') { const END_DAY = new Date(dateA) const START_DAY = new Date(DateB) let calculatedDateBy let returnDateDiff if (type === 'month') { const startMonth = START_DAY.getMonth() const endMonth = END_DAY.getMonth() calculatedDateBy = startMonth - endMonth returnDateDiff = Math.abs( calculatedDateBy + 12 * (START_DAY.getFullYear() - END_DAY.getFullYear()) ) } else { calculatedDateBy = Math.abs(START_DAY - END_DAY) returnDateDiff = Math.ceil(calculatedDateBy / (1000 * 60 * 60 * 24)) } const out = document.getElementById('output') out.innerText = returnDateDiff return returnDateDiff } // Gets number of days from given dates /* getDateDifference('2022-03-31','2022-04-08','day') */ // Get number of months from given dates getDateDifference('2021-12-02','2022-04-08','month') <div id="output"> </div>
function monthDiff(d1, d2) {
var months, d1day, d2day, d1new, d2new, diffdate,d2month,d2year,d1maxday,d2maxday;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
months += d2.getMonth();
months = (months <= 0 ? 0 : months);
d1day = d1.getDate();
d2day = d2.getDate();
if(d1day > d2day)
{
d2month = d2.getMonth();
d2year = d2.getFullYear();
d1new = new Date(d2year, d2month-1, d1day,0,0,0,0);
var timeDiff = Math.abs(d2.getTime() - d1new.getTime());
diffdate = Math.abs(Math.ceil(timeDiff / (1000 * 3600 * 24)));
d1new = new Date(d2year, d2month, 1,0,0,0,0);
d1new.setDate(d1new.getDate()-1);
d1maxday = d1new.getDate();
months += diffdate / d1maxday;
}
else
{
if(!(d1.getMonth() == d2.getMonth() && d1.getFullYear() == d2.getFullYear()))
{
months += 1;
}
diffdate = d2day - d1day + 1;
d2month = d2.getMonth();
d2year = d2.getFullYear();
d2new = new Date(d2year, d2month + 1, 1, 0, 0, 0, 0);
d2new.setDate(d2new.getDate()-1);
d2maxday = d2new.getDate();
months += diffdate / d2maxday;
}
return months;
}
任何值连同它的绝对值一起返回。
function differenceInMonths(firstDate, secondDate) {
if (firstDate > secondDate) [firstDate, secondDate] = [secondDate, firstDate];
let diffMonths = (secondDate.getFullYear() - firstDate.getFullYear()) * 12;
diffMonths -= firstDate.getMonth();
diffMonths += secondDate.getMonth();
return diffMonths;
}
“月差数”的定义有很多不同的解释。: -)
您可以从JavaScript date对象中获取年、月和月中的日期。根据您要查找的信息,您可以使用这些信息来计算出两个时间点之间有多少个月。
例如,off- cuff:
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth();
months += d2.getMonth();
return months <= 0 ? 0 : months;
}
function monthDiff(d1, d2) { var months; months = (d2.getFullYear() - d1.getFullYear()) * 12; months -= d1.getMonth(); months += d2.getMonth(); return months <= 0 ? 0 : months; } function test(d1, d2) { var diff = monthDiff(d1, d2); console.log( d1.toISOString().substring(0, 10), "to", d2.toISOString().substring(0, 10), ":", diff ); } test( new Date(2008, 10, 4), // November 4th, 2008 new Date(2010, 2, 12) // March 12th, 2010 ); // Result: 16 test( new Date(2010, 0, 1), // January 1st, 2010 new Date(2010, 2, 12) // March 12th, 2010 ); // Result: 2 test( new Date(2010, 1, 1), // February 1st, 2010 new Date(2010, 2, 12) // March 12th, 2010 ); // Result: 1
(请注意,JavaScript中的月份值以0 = January开始。)
包括上面的小数月份要复杂得多,因为典型的2月份的三天(~10.714%)比8月份的三天(~9.677%)占这个月的比例要大得多,当然,甚至2月份也是一个移动的目标,这取决于它是否是闰年。
还有一些可供JavaScript使用的日期和时间库,可能会使这类事情变得更容易。
注:上面曾经有一个+ 1,这里:
months = (d2.getFullYear() - d1.getFullYear()) * 12;
months -= d1.getMonth() + 1;
// −−−−−−−−−−−−−−−−−−−−^^^^
months += d2.getMonth();
因为一开始我说
...这将找出两个日期之间有多少个完整的月份,不包括部分月份(例如,不包括每个日期所在的月份)。
我删除它有两个原因:
结果发现,不计算部分月份并不是很多(大多数?)人想得到的答案,所以我认为我应该把它们分开。 即使按照这个定义,它也并不总是有效。: - d(抱歉。)
下面是另一种更少循环的方法:
calculateTotalMonthsDifference = function(firstDate, secondDate) {
var fm = firstDate.getMonth();
var fy = firstDate.getFullYear();
var sm = secondDate.getMonth();
var sy = secondDate.getFullYear();
var months = Math.abs(((fy - sy) * 12) + fm - sm);
var firstBefore = firstDate > secondDate;
firstDate.setFullYear(sy);
firstDate.setMonth(sm);
firstBefore ? firstDate < secondDate ? months-- : "" : secondDate < firstDate ? months-- : "";
return months;
}