Try something like this

dateLimit = (curDate, limit) => {
    offset  = curDate.getDate() + limit
    return new Date( curDate.setDate( offset) )
}

currDate could be any date

limit could be the difference in number of day (positive for future and negative for past)

Using Modern JavaScript function syntax

const getDaysPastDate = (daysBefore, date = new Date) => new Date(date - (1000 * 60 * 60 * 24 * daysBefore));

console.log(getDaysPastDate(1)); // yesterday

If you want to both subtract a number of days and format your date in a human readable format, you should consider creating a custom DateHelper object that looks something like this :

var DateHelper = {
    addDays : function(aDate, numberOfDays) {
        aDate.setDate(aDate.getDate() + numberOfDays); // Add numberOfDays
        return aDate;                                  // Return the date
    },
    format : function format(date) {
        return [
           ("0" + date.getDate()).slice(-2),           // Get day and pad it with zeroes
           ("0" + (date.getMonth()+1)).slice(-2),      // Get month and pad it with zeroes
           date.getFullYear()                          // Get full year
        ].join('/');                                   // Glue the pieces together
    }
}

// With this helper, you can now just use one line of readable code to :
// ---------------------------------------------------------------------
// 1. Get the current date
// 2. Subtract 5 days
// 3. Format it
// 4. Output it
// ---------------------------------------------------------------------
document.body.innerHTML = DateHelper.format(DateHelper.addDays(new Date(), -5));

(see also this Fiddle)

I get good mileage out of date.js:

http://www.datejs.com/

d = new Date();
d.add(-10).days();  // subtract 10 days

Nice!

Website includes this beauty:

Datejs doesn’t just parse strings, it slices them cleanly in two

See the following code, subtract the days from the current date. Also, set the month according to substracted date.

var today = new Date();
var substract_no_of_days = 25;

today.setTime(today.getTime() - substract_no_of_days* 24 * 60 * 60 * 1000);
var substracted_date = (today.getMonth()+1) + "/" +today.getDate() + "/" + today.getFullYear();

alert(substracted_date);

When setting the date, the date converts to milliseconds, so you need to convert it back to a date:

This method also take into consideration, new year change etc.

function addDays( date, days ) {
    var dateInMs = date.setDate(date.getDate() - days);
    return new Date(dateInMs);
}

var date_from = new Date();
var date_to = addDays( new Date(), parseInt(days) );

for me all the combinations worked fine with below code snipplet , the snippet is for Angular-2 implementation , if you need to add days , pass positive numberofDays , if you need to substract pass negative numberofDays

function addSubstractDays(date: Date, numberofDays: number): Date {
let d = new Date(date);
return new Date(
    d.getFullYear(),
    d.getMonth(),
    (d.getDate() + numberofDays)
);
}

最佳答案导致我的代码出现错误，该错误将在本月的第一天设置当前月份的将来日期。这就是我所做的

curDate = new Date(); // Took current date as an example
prvDate = new Date(0); // Date set to epoch 0
prvDate.setUTCMilliseconds((curDate - (5 * 24 * 60 * 60 * 1000))); //Set epoch time

I like the following because it is one line. Not perfect with DST changes but usually good enough for my needs.

var fiveDaysAgo = new Date(new Date() - (1000*60*60*24*5));

我创建了一个用于日期操作的函数。您可以添加或减去任何天数，小时数，分钟数。

function dateManipulation(date, days, hrs, mins, operator) {
   date = new Date(date);
   if (operator == "-") {
      var durationInMs = (((24 * days) * 60) + (hrs * 60) + mins) * 60000;
      var newDate = new Date(date.getTime() - durationInMs);
   } else {
      var durationInMs = (((24 * days) * 60) + (hrs * 60) + mins) * 60000;
      var newDate = new Date(date.getTime() + durationInMs);
   }
   return newDate;
 }

现在，通过传递参数来调用此函数。例如，这是一个函数调用，用于获取从今天起3天之前的日期。

var today = new Date();
var newDate = dateManipulation(today, 3, 0, 0, "-");

无需使用第二个变量，您可以将x替换为x天：

let d=new Date(new Date().getTime() - (7 * 24 * 60 * 60 * 1000))

我发现getDate（）/ setDate（）方法存在一个问题，即它太容易将所有内容转换为毫秒，并且有时语法难以理解。

相反，我想解决一个事实，即1天= 86,400,000毫秒。

因此，对于您的特定问题：

today = new Date()
days = 86400000 //number of milliseconds in a day
fiveDaysAgo = new Date(today - (5*days))

奇迹般有效。

我一直在使用这种方法进行30/60/365天的滚动计算。

您可以轻松地对此进行推断，以创建几个月，几年等的时间单位。

一些现有的解决方案已经接近，但与我想要的不完全相同。此函数可使用正值或负值，并处理边界情况。

function addDays(date, days) {
    return new Date(
        date.getFullYear(),
        date.getMonth(),
        date.getDate() + days,
        date.getHours(),
        date.getMinutes(),
        date.getSeconds(),
        date.getMilliseconds()
    );
}

我喜欢以毫秒为单位进行数学运算。所以用Date.now()

var newDate = Date.now() + -5*24*3600*1000; // date 5 days ago in milliseconds

如果您喜欢它的格式

new Date(newDate).toString(); // or .toUTCString or .toISOString ...

注意：Date.now()在较旧的浏览器（例如我认为的IE8）中不起作用。Polyfill在这里。

2015年6月更新

@socketpair指出了我的草率。正如他/她所说：“由于时区规则，一年中的某天有23个小时，有25个小时”。

进一步来说，如果您想计算夏令时更改的时区中的5天前的本地日，并且上面的答案存在夏令时错误，并且您

• 假设（错误地）Date.now()为您提供了当前的本地时间，或者
• 用途.toString()，它返回本地日期，因此与Date.now()UTC中的基准日期不兼容。

但是，如果您全部使用UTC进行数学运算，则可以使用它，例如

答：您希望UTC日期从NOW（UTC）5天前开始

var newDate = Date.now() + -5*24*3600*1000; // date 5 days ago in milliseconds UTC
new Date(newDate).toUTCString(); // or .toISOString(), BUT NOT toString

B.您使用UTC基准日期（不是“现在”）开始，使用 Date.UTC()

newDate = new Date(Date.UTC(2015, 3, 1)).getTime() + -5*24*3600000;
new Date(newDate).toUTCString(); // or .toISOString BUT NOT toString

我注意到getDays + X在日期/月份范围内不起作用。只要您的日期不早于1970，就可以使用getTime。

var todayDate = new Date(), weekDate = new Date();
weekDate.setTime(todayDate.getTime()-(7*24*3600000));

我为Date创建了此原型，以便可以传递负值减去天数，传递正值添加天数。

if(!Date.prototype.adjustDate){
    Date.prototype.adjustDate = function(days){
        var date;

        days = days || 0;

        if(days === 0){
            date = new Date( this.getTime() );
        } else if(days > 0) {
            date = new Date( this.getTime() );

            date.setDate(date.getDate() + days);
        } else {
            date = new Date(
                this.getFullYear(),
                this.getMonth(),
                this.getDate() - Math.abs(days),
                this.getHours(),
                this.getMinutes(),
                this.getSeconds(),
                this.getMilliseconds()
            );
        }

        this.setTime(date.getTime());

        return this;
    };
}

因此，要使用它，我可以简单地写：

var date_subtract = new Date().adjustDate(-4),
    date_add = new Date().adjustDate(4);

它是这样的：

var d = new Date(); // today!
var x = 5; // go back 5 days!
d.setDate(d.getDate() - x);

var dateOffset = (24*60*60*1000) * 5; //5 days
var myDate = new Date();
myDate.setTime(myDate.getTime() - dateOffset);

如果您要在整个Web应用程序中执行很多头痛检查日期操作，DateJS将使您的生活更加轻松：

http://simonwillison.net/2007/Dec/3/datejs/