我觉得这里的许多答案都缺乏创造性的成分，这是时间旅行计算非常需要的。我提出了30分钟时间翻译的解决方案。

(这里jsfiddle)

function fluxCapacitor(n) {
    var delta,sigma=0,beta="ge";
    (function(K,z){

        (function(a,b,c){
            beta=beta+"tT";
            switch(b.shift()) {
                case'3':return z('0',a,c,b.shift(),1);
                case'0':return z('3',a,c,b.pop());
                case'5':return z('2',a,c,b[0],1);
                case'1':return z('4',a,c,b.shift());
                case'2':return z('5',a,c,b.pop());
                case'4':return z('1',a,c,b.pop(),1);
            }
        })(K.pop(),K.pop().split(''),K.pop());
    })(n.toString().split(':'),function(b,a,c,b1,gamma){
       delta=[c,b+b1,a];sigma+=gamma?3600000:0; 
       beta=beta+"im";
    });
    beta=beta+"e";
    return new Date (sigma+(new Date( delta.join(':')))[beta]());
}

我知道这个话题太老了。但是我非常确定有一些开发人员仍然需要这个，所以我为您制作了这个简单的脚本。 希望你喜欢!

你好，现在是2020年，我做了一些修改，希望现在能更好地帮助你!

你好，现在是2022年，我又回来解决了一些问题，并为方法和函数提供了更好的命名。

function addTimeToDate(addedTime, date){ let generatedTime = date.getTime(); if(addedTime.seconds) generatedTime += 1000 * addedTime.seconds; //check for additional seconds if(addedTime.minutes) generatedTime += 1000* 60 * addedTime.minutes;//check for additional minutes if(addedTime.hours) generatedTime += 1000 * 60 * 60 * addedTime.hours;//check for additional hours return new Date(generatedTime); } Date.prototype.addTime = function(addedTime){ return addTimeToDate(addedTime, this); } let futureDate = new Date().addTime({ hours: 16, //Adding one hour minutes: 45, //Adding fourty five minutes seconds: 0 //Adding 0 seconds return to not adding any second so we can remove it. }); <button onclick="console.log(futureDate)">Travel to the future</button>

对于像我这样的懒人:

Kip的答案(从上面)在coffeescript中，使用“enum”，并对同一对象进行操作:

Date.UNIT =
  YEAR: 0
  QUARTER: 1
  MONTH: 2
  WEEK: 3
  DAY: 4
  HOUR: 5
  MINUTE: 6
  SECOND: 7
Date::add = (unit, quantity) ->
  switch unit
    when Date.UNIT.YEAR then @setFullYear(@getFullYear() + quantity)
    when Date.UNIT.QUARTER then @setMonth(@getMonth() + (3 * quantity))
    when Date.UNIT.MONTH then @setMonth(@getMonth() + quantity)
    when Date.UNIT.WEEK then @setDate(@getDate() + (7 * quantity))
    when Date.UNIT.DAY then @setDate(@getDate() + quantity)
    when Date.UNIT.HOUR then @setTime(@getTime() + (3600000 * quantity))
    when Date.UNIT.MINUTE then @setTime(@getTime() + (60000 * quantity))
    when Date.UNIT.SECOND then @setTime(@getTime() + (1000 * quantity))
    else throw new Error "Unrecognized unit provided"
  @ # for chaining

我总是创建7个函数，在JS中使用date: addSeconds, addMinutes, addHours, addDays, addWeeks, addMonths, addYears。

你可以在这里看到一个例子:http://jsfiddle.net/tiagoajacobi/YHA8x/

使用方法:

var now = new Date();
console.log(now.addMinutes(30));
console.log(now.addWeeks(3));

这些是函数:

Date.prototype.addSeconds = function(seconds) {
  this.setSeconds(this.getSeconds() + seconds);
  return this;
};

Date.prototype.addMinutes = function(minutes) {
  this.setMinutes(this.getMinutes() + minutes);
  return this;
};

Date.prototype.addHours = function(hours) {
  this.setHours(this.getHours() + hours);
  return this;
};

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

Date.prototype.addWeeks = function(weeks) {
  this.addDays(weeks*7);
  return this;
};

Date.prototype.addMonths = function (months) {
  var dt = this.getDate();
  this.setMonth(this.getMonth() + months);
  var currDt = this.getDate();
  if (dt !== currDt) {  
    this.addDays(-currDt);
  }
  return this;
};

Date.prototype.addYears = function(years) {
  var dt = this.getDate();
  this.setFullYear(this.getFullYear() + years);
  var currDt = this.getDate();
  if (dt !== currDt) {  
    this.addDays(-currDt);
  }
  return this;
};

你可以这样做:

let 30tyminutes = 30 * 60 * 1000;//将30分钟转换为毫秒 let date1 = new Date(); let date2 = new Date(date1.getTime() + 30tyminutes); console.log (date1); console.log (date2);

简单地，你可以在momnet库中使用这段代码:

console.log(moment(moment()).add(30,"minutes").format('MM/DD/YYYY hh:mm:ss'));