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
下面是一个函数,它精确地提供了两个日期之间的月数。 默认行为只计算整个月份,例如3个月和1天将导致3个月的差异。您可以通过将roundUpFractionalMonths参数设置为true来防止这种情况,因此3个月和1天的差异将返回为4个月。
上面公认的答案(T.J.克劳德的答案)是不准确的,它有时会返回错误的值。
例如,monthDiff(new Date('Jul 01, 2015'), new Date('Aug 05, 2015'))返回0,这显然是错误的。正确的差值是1个月或2个月。
这是我写的函数:
function getMonthsBetween(date1,date2,roundUpFractionalMonths)
{
//Months will be calculated between start and end dates.
//Make sure start date is less than end date.
//But remember if the difference should be negative.
var startDate=date1;
var endDate=date2;
var inverse=false;
if(date1>date2)
{
startDate=date2;
endDate=date1;
inverse=true;
}
//Calculate the differences between the start and end dates
var yearsDifference=endDate.getFullYear()-startDate.getFullYear();
var monthsDifference=endDate.getMonth()-startDate.getMonth();
var daysDifference=endDate.getDate()-startDate.getDate();
var monthCorrection=0;
//If roundUpFractionalMonths is true, check if an extra month needs to be added from rounding up.
//The difference is done by ceiling (round up), e.g. 3 months and 1 day will be 4 months.
if(roundUpFractionalMonths===true && daysDifference>0)
{
monthCorrection=1;
}
//If the day difference between the 2 months is negative, the last month is not a whole month.
else if(roundUpFractionalMonths!==true && daysDifference<0)
{
monthCorrection=-1;
}
return (inverse?-1:1)*(yearsDifference*12+monthsDifference+monthCorrection);
};
其他回答
这是我能找到的最简单的解。这将直接返回月数。尽管,它总是给出一个绝对值。
new Date(new Date(d2) - new Date(d1)).getMonth();
对于非绝对值,您可以使用以下解决方案:
function diff_months(startDate, endDate) {
let diff = new Date( new Date(endDate) - new Date(startDate) ).getMonth();
return endDate >= startDate ? diff : -diff;
}
JavaScript中两个日期的月份差异:
start_date = new Date(year, month, day); //Create start date object by passing appropiate argument
end_date = new Date(new Date(year, month, day)
从start_date到end_date的总月份:
total_months = (end_date.getFullYear() - start_date.getFullYear())*12 + (end_date.getMonth() - start_date.getMonth())
anyVar = (((DisplayTo.getFullYear() * 12) + DisplayTo.getMonth()) - ((DisplayFrom.getFullYear() * 12) + DisplayFrom.getMonth()));
你也可以考虑这个解决方案,这个函数返回整数或数字形式的月差
将开始日期作为第一个或最后一个参数传递是容错的。这意味着,函数仍然会返回相同的值。
const diffInMonths = (end, start) => { var timeDiff = Math.abs(end.getTime() - start.getTime()); 返回数学。round(timeDiff / (2e3 * 3600 * 365.25)); } const result = diffInMonths(new Date(2015, 3,28), new Date(2010, 1,25)); //显示月差值为整数/数字 console.log(结果);
它还计算天数并以月为单位进行转换。
function monthDiff(d1, d2) {
var months;
months = (d2.getFullYear() - d1.getFullYear()) * 12; //calculates months between two years
months -= d1.getMonth() + 1;
months += d2.getMonth(); //calculates number of complete months between two months
day1 = 30-d1.getDate();
day2 = day1 + d2.getDate();
months += parseInt(day2/30); //calculates no of complete months lie between two dates
return months <= 0 ? 0 : months;
}
monthDiff(
new Date(2017, 8, 8), // Aug 8th, 2017 (d1)
new Date(2017, 12, 12) // Dec 12th, 2017 (d2)
);
//return value will be 4 months