如何在今天的日期添加天数？ [重复]-Java 学习之路

257

这个问题在这里已有答案：

Add days to JavaScript Date 37个答案

我需要能够使用jQuery在今天的日期添加1,2,5或10天 .

16 回答

你可以使用JavaScript，不需要jQuery：

var someDate = new Date();
var numberOfDaysToAdd = 6;
someDate.setDate(someDate.getDate() + numberOfDaysToAdd);

格式为 dd/mm/yyyy ：

var dd = someDate.getDate();
var mm = someDate.getMonth() + 1;
var y = someDate.getFullYear();

var someFormattedDate = dd + '/'+ mm + '/'+ y;

回复于 2024-05-03T07:18:20+08:00

您可以像这样扩展javascript Date对象

Date.prototype.addDays = function(days) {
    this.setDate(this.getDate() + parseInt(days));
    return this;
};

并在您的javascript代码中，您可以调用

var currentDate = new Date();
// to add 4 days to current date
currentDate.addDays(4);

回复于 2024-05-03T07:18:20+08:00

为什么不简单地使用

function addDays(theDate, days) {
    return new Date(theDate.getTime() + days*24*60*60*1000);
}

var newDate = addDays(new Date(), 5);

或-5删除5天

回复于 2024-05-03T07:18:20+08:00

0
这是5天：
```
var myDate = new Date(new Date().getTime()+(5*24*60*60*1000));
```
你不需要JQuery，你可以在JavaScript中完成它，希望你能得到它 .
回复于 2024-05-03T07:18:20+08:00
4
Moment.js

从here安装 moment.js .

npm ： $ npm i --save moment

Bower ：$ bower install --save moment

下一个，
```
var date = moment()
            .add(2,'d') //replace 2 with number of days you want to add
            .toDate(); //convert it to a Javascript Date Object if you like
```
链接参考：http://momentjs.com/docs/#/manipulating/add/

Moment.js 是一个令人惊叹的Javascript库，可以在 40kb 管理Date对象和极轻的权重 .

祝好运 .
回复于 2024-05-03T07:18:20+08:00
0
来自Krishna Chytanya的原型解决方案非常好，但需要一个小的但重要的改进 . 必须将days param解析为Integer，以避免在days为类似“1”的字符串时进行奇怪的计算 . （我需要几个小时才能找到，我的申请出了什么问题 . ）
```
Date.prototype.addDays = function(days) {
    this.setDate(this.getDate() + parseInt(days));
    return this;
};
```
即使您不使用此原型功能： Always be sure to have an Integer when using setDate().
回复于 2024-05-03T07:18:20+08:00
4
这里接受的答案给了我不可预知的结果，有时奇怪地增加了几个月和几年 .

我能找到的最可靠的方法是在这里找到Add days to Javascript Date object, and also increment month
```
var dayOffset = 20;
var millisecondOffset = dayOffset * 24 * 60 * 60 * 1000;
december.setTime(december.getTime() + millisecondOffset);
```
编辑：虽然它对一些人有用，但我认为这不完全正确 . 我建议使用更受欢迎的答案或使用像http://momentjs.com/这样的东西
回复于 2024-05-03T07:18:20+08:00

432

纯JS解决方案，日期格式为YYYY-mm-dd格式

var someDate = new Date('2014-05-14');
someDate.setDate(someDate.getDate() + 15); //number  of days to add, e.x. 15 days
var dateFormated = someDate.toISOString().substr(0,10);

回复于 2024-05-03T07:18:20+08:00

Date.prototype.addDays = function(days)
{
    var dat = new Date(this.valueOf() + days * 24 * 60 * 60 * 1000 );
    return dat;
}

回复于 2024-05-03T07:18:20+08:00

27
如果不需要日期中的时间，那么您只需使用日期对象的方法来提取月，年和日，并向日部分添加“n”天数 .
```
var n=5; //number of days to add. 
var today=new Date(); //Today's Date
var requiredDate=new Date(today.getFullYear(),today.getMonth(),today.getDate()+n)
```
参考：Mozilla Javascript GetDate

编辑：参考：Mozilla JavaScript Date
回复于 2024-05-03T07:18:20+08:00
68

您可以使用此库“Datejs开源JavaScript日期库” .

回复于 2024-05-03T07:18:20+08:00

我发现这是javascript的痛苦 . 看看这个帮助我的链接 . 你有没有想过扩展日期对象 .

http://pristinecoder.com/Blog/post/javascript-formatting-date-in-javascript

/*
 * Date Format 1.2.3
 * (c) 2007-2009 Steven Levithan <stevenlevithan.com>
 * MIT license
 *
 * Includes enhancements by Scott Trenda <scott.trenda.net>
 * and Kris Kowal <cixar.com/~kris.kowal/>
 *
 * Accepts a date, a mask, or a date and a mask.
 * Returns a formatted version of the given date.
 * The date defaults to the current date/time.
 * The mask defaults to dateFormat.masks.default.
 */

var dateFormat = function () {
    var token = /d{1,4}|m{1,4}|yy(?:yy)?|([HhMsTt])\1?|[LloSZ]|"[^"]*"|'[^']*'/g,
        timezone = /\b(?:[PMCEA][SDP]T|(?:Pacific|Mountain|Central|Eastern|Atlantic) (?:Standard|Daylight|Prevailing) Time|(?:GMT|UTC)(?:[-+]\d{4})?)\b/g,
        timezoneClip = /[^-+\dA-Z]/g,
        pad = function (val, len) {
            val = String(val);
            len = len || 2;
            while (val.length < len) val = "0" + val;
            return val;
        };

    // Regexes and supporting functions are cached through closure
    return function (date, mask, utc) {
        var dF = dateFormat;

        // You can't provide utc if you skip other args (use the "UTC:" mask prefix)
        if (arguments.length == 1 && Object.prototype.toString.call(date) == "[object String]" && !/\d/.test(date)) {
            mask = date;
            date = undefined;
        }

        // Passing date through Date applies Date.parse, if necessary
        date = date ? new Date(date) : new Date;
        if (isNaN(date)) throw SyntaxError("invalid date");

        mask = String(dF.masks[mask] || mask || dF.masks["default"]);

        // Allow setting the utc argument via the mask
        if (mask.slice(0, 4) == "UTC:") {
            mask = mask.slice(4);
            utc = true;
        }

        var _ = utc ? "getUTC" : "get",
            d = date[_ + "Date"](),
            D = date[_ + "Day"](),
            m = date[_ + "Month"](),
            y = date[_ + "FullYear"](),
            H = date[_ + "Hours"](),
            M = date[_ + "Minutes"](),
            s = date[_ + "Seconds"](),
            L = date[_ + "Milliseconds"](),
            o = utc ? 0 : date.getTimezoneOffset(),
            flags = {
                d:    d,
                dd:   pad(d),
                ddd:  dF.i18n.dayNames[D],
                dddd: dF.i18n.dayNames[D + 7],
                m:    m + 1,
                mm:   pad(m + 1),
                mmm:  dF.i18n.monthNames[m],
                mmmm: dF.i18n.monthNames[m + 12],
                yy:   String(y).slice(2),
                yyyy: y,
                h:    H % 12 || 12,
                hh:   pad(H % 12 || 12),
                H:    H,
                HH:   pad(H),
                M:    M,
                MM:   pad(M),
                s:    s,
                ss:   pad(s),
                l:    pad(L, 3),
                L:    pad(L > 99 ? Math.round(L / 10) : L),
                t:    H < 12 ? "a"  : "p",
                tt:   H < 12 ? "am" : "pm",
                T:    H < 12 ? "A"  : "P",
                TT:   H < 12 ? "AM" : "PM",
                Z:    utc ? "UTC" : (String(date).match(timezone) || [""]).pop().replace(timezoneClip, ""),
                o:    (o > 0 ? "-" : "+") + pad(Math.floor(Math.abs(o) / 60) * 100 + Math.abs(o) % 60, 4),
                S:    ["th", "st", "nd", "rd"][d % 10 > 3 ? 0 : (d % 100 - d % 10 != 10) * d % 10]
            };

        return mask.replace(token, function ($0) {
            return $0 in flags ? flags[$0] : $0.slice(1, $0.length - 1);
        });
    };
}();

// Some common format strings
dateFormat.masks = {
    "default":      "ddd mmm dd yyyy HH:MM:ss",
    shortDate:      "m/d/yy",
    mediumDate:     "mmm d, yyyy",
    longDate:       "mmmm d, yyyy",
    fullDate:       "dddd, mmmm d, yyyy",
    shortTime:      "h:MM TT",
    mediumTime:     "h:MM:ss TT",
    longTime:       "h:MM:ss TT Z",
    isoDate:        "yyyy-mm-dd",
    isoTime:        "HH:MM:ss",
    isoDateTime:    "yyyy-mm-dd'T'HH:MM:ss",
    isoUtcDateTime: "UTC:yyyy-mm-dd'T'HH:MM:ss'Z'"
};

// Internationalization strings
dateFormat.i18n = {
    dayNames: [
        "Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat",
        "Sunday", "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday"
    ],
    monthNames: [
        "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec",
        "January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"
    ]
};

// For convenience...
Date.prototype.format = function (mask, utc) {
    return dateFormat(this, mask, utc);
};

回复于 2024-05-03T07:18:20+08:00

function addDays(n){
    var t = new Date();
    t.setDate(t.getDate() + n); 
    var month = "0"+(t.getMonth()+1);
    var date = "0"+t.getDate();
    month = month.slice(-2);
    date = date.slice(-2);
     var date = date +"/"+month +"/"+t.getFullYear();
    alert(date);
}

addDays(5);

回复于 2024-05-03T07:18:20+08:00

我发现当你使用 new Date(nYear, nMonth, nDate); 时，JavaScript可以返回正确的日期 . 使用它时，尝试查看 dDate 变量的结果：

var dDate = new Date(2012, 0, 34); // the result is 3 Feb 2012

我有一个SkipDate函数可以分享：

function DaysOfMonth(nYear, nMonth) {
        switch (nMonth) {
            case 0:     // January
                return 31; break;
            case 1:     // February
                if ((nYear % 4) == 0) {
                    return 29;
                }
                else {
                    return 28;
                };
                break;
            case 2:     // March
                return 31; break;
            case 3:     // April
                return 30; break;
            case 4:     // May
                return 31; break;
            case 5:     // June
                return 30; break;
            case 6:     // July
                return 31; break;
            case 7:     // August
                return 31; break;
            case 8:     // September
                return 30; break;
            case 9:     // October
                return 31; break;
            case 10:     // November
                return 30; break;
            case 11:     // December
                return 31; break;
        }
    };

    function SkipDate(dDate, skipDays) {
        var nYear = dDate.getFullYear();
        var nMonth = dDate.getMonth();
        var nDate = dDate.getDate();
        var remainDays = skipDays;
        var dRunDate = dDate;

        while (remainDays > 0) {
            remainDays_month = DaysOfMonth(nYear, nMonth) - nDate;
            if (remainDays > remainDays_month) {
                remainDays = remainDays - remainDays_month - 1;
                nDate = 1;
                if (nMonth < 11) { nMonth = nMonth + 1; }
                else {
                    nMonth = 0;
                    nYear = nYear + 1;
                };
            }
            else {
                nDate = nDate + remainDays;
                remainDays = 0;
            };
            dRunDate = Date(nYear, nMonth, nDate);
        }
        return new Date(nYear, nMonth, nDate);
    };

回复于 2024-05-03T07:18:20+08:00

这是一个适合我的解决方案 .

function calduedate(ndays){

    var newdt = new Date(); var chrday; var chrmnth;
    newdt.setDate(newdt.getDate() + parseInt(ndays));

    var newdate = newdt.getFullYear();
    if(newdt.getMonth() < 10){
        newdate = newdate+'-'+'0'+newdt.getMonth();
    }else{
        newdate = newdate+'-'+newdt.getMonth();
    }
    if(newdt.getDate() < 10){
        newdate = newdate+'-'+'0'+newdt.getDate();
    }else{
        newdate = newdate+'-'+newdt.getDate();
    }

    alert("newdate="+newdate);

}

回复于 2024-05-03T07:18:20+08:00

50
你可以尝试这个，不需要JQuery：timeSolver.js

例如，今天添加5天：
```
var newDay = timeSolver.add(new Date(),5,"day");
```
你也可以按小时，月等添加 . 请参阅更多信息 .
回复于 2024-05-03T07:18:20+08:00

如何在今天的日期添加天数？ [重复]

16 回答

相关问题