如何使用JavaScript将日期添加到当前日期?JavaScript是否有像.NET的AddDay()那样的内置函数?


当前回答

我昨晚创建了这些扩展:可以传递正值或负值;

例子:

var someDate = new Date();
var expirationDate = someDate.addDays(10);
var previous = someDate.addDays(-5);


Date.prototype.addDays = function (num) {
    var value = this.valueOf();
    value += 86400000 * num;
    return new Date(value);
}

Date.prototype.addSeconds = function (num) {
    var value = this.valueOf();
    value += 1000 * num;
    return new Date(value);
}

Date.prototype.addMinutes = function (num) {
    var value = this.valueOf();
    value += 60000 * num;
    return new Date(value);
}

Date.prototype.addHours = function (num) {
    var value = this.valueOf();
    value += 3600000 * num;
    return new Date(value);
}

Date.prototype.addMonths = function (num) {
    var value = new Date(this.valueOf());

    var mo = this.getMonth();
    var yr = this.getYear();

    mo = (mo + num) % 12;
    if (0 > mo) {
        yr += (this.getMonth() + num - mo - 12) / 12;
        mo += 12;
    }
    else
        yr += ((this.getMonth() + num - mo) / 12);

    value.setMonth(mo);
    value.setYear(yr);
    return value;
}

其他回答

    //the_day is 2013-12-31
    var the_day = Date.UTC(2013, 11, 31); 
    // Now, the_day will be "1388448000000" in UTC+8; 
    var the_next_day = new Date(the_day + 24 * 60 * 60 * 1000);
    // Now, the_next_day will be "Wed Jan 01 2014 08:00:00 GMT+0800"

Try

var someDate = new Date();
var duration = 2; //In Days
someDate.setTime(someDate.getTime() +  (duration * 24 * 60 * 60 * 1000));

使用setDate()添加日期并不能解决您的问题,请尝试在2月份添加一些天,如果您尝试在其中添加新的天,则不会产生您预期的结果。

我的测试示例可以在日期对象的同一实例中执行减号。

Date.prototype.reset=函数(){let newDate=新日期(this.timeStamp)this.setFullYear(newDate.getFullYear)this.setMonth(newDate.getMonth())this.setDate(newDate.getDate())this.setHours(newDate.getHours())this.set分钟(newDate.getMinutes())this.setSeconds(newDate.getSeconds())this.set毫秒(newDate.getMilliseconds())}Date.prototype.addDays=函数(天){this.timeStamp=此[Symbol.toPrimitive]('编号')let daysInMiliseconds=(天*(1000*60*60*24))this.timeStamp=this.timeStamp+天毫秒this.reset()}Date.prototype.minusDays=函数(天){this.timeStamp=此[Symbol.toPrimitive]('编号')let daysInMiliseconds=(天*(1000*60*60*24))如果(daysInMiliseconds<=this.timeStamp){this.timeStamp=this.timeStamp-天毫秒this.reset()}}var temp=新日期(Date.now())//从现在开始console.log(temp.toDateString())临时添加天数(31)console.log(temp.toDateString())温度-天(5)console.log(temp.toDateString())

我昨晚创建了这些扩展:可以传递正值或负值;

例子:

var someDate = new Date();
var expirationDate = someDate.addDays(10);
var previous = someDate.addDays(-5);


Date.prototype.addDays = function (num) {
    var value = this.valueOf();
    value += 86400000 * num;
    return new Date(value);
}

Date.prototype.addSeconds = function (num) {
    var value = this.valueOf();
    value += 1000 * num;
    return new Date(value);
}

Date.prototype.addMinutes = function (num) {
    var value = this.valueOf();
    value += 60000 * num;
    return new Date(value);
}

Date.prototype.addHours = function (num) {
    var value = this.valueOf();
    value += 3600000 * num;
    return new Date(value);
}

Date.prototype.addMonths = function (num) {
    var value = new Date(this.valueOf());

    var mo = this.getMonth();
    var yr = this.getYear();

    mo = (mo + num) % 12;
    if (0 > mo) {
        yr += (this.getMonth() + num - mo - 12) / 12;
        mo += 12;
    }
    else
        yr += ((this.getMonth() + num - mo) / 12);

    value.setMonth(mo);
    value.setYear(yr);
    return value;
}

我真不敢相信,5年后,这条线索中没有捷径可走!SO:要获得一天中相同的时间,而不考虑夏季的干扰:

Date.prototype.addDays = function(days)
    {
    var dat = new Date( this.valueOf() )

    var hour1 = dat.getHours()
    dat.setTime( dat.getTime() + days * 86400000) // 24*60*60*1000 = 24 hours
    var hour2 = dat.getHours()

    if (hour1 != hour2) // summertime occured +/- a WHOLE number of hours thank god!
        dat.setTime( dat.getTime() + (hour1 - hour2) * 3600000) // 60*60*1000 = 1 hour

    return dat
or
    this.setTime( dat.getTime() ) // to modify the object directly
    }

那里完成!