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 calcualteMonthYr(){
    var fromDate =new Date($('#txtDurationFrom2').val()); //date picker (text fields)
    var toDate = new Date($('#txtDurationTo2').val());

var months=0;
        months = (toDate.getFullYear() - fromDate.getFullYear()) * 12;
        months -= fromDate.getMonth();
        months += toDate.getMonth();
            if (toDate.getDate() < fromDate.getDate()){
                months--;
            }
    $('#txtTimePeriod2').val(months);
}

如果你不考虑每月的日期,这是迄今为止最简单的解决方案

function monthDiff(dateTo, dateTo) 回来了 (12 * (dateTo.getFullYear) - dateyear .getFullYear() 的 /和 控制台(新日期(2000年,01年),新日期(2000年,02年)// 1 控制台,log(monthDiff, new Date(1999, 02), new Date(2000, 02)) // 12年 控制台.log(monthDiff,新日期(2009年,11年),新日期(2010年,0)// 1

注意,月份索引是基于0的。这意味着一月= 0,十二月= 11。

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())
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;
}