我想获得一个日期对象,它比另一个日期对象晚30分钟。我如何用JavaScript做到这一点?


当前回答

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);

其他回答

以下是ES6版本:

let getTimeAfter30Mins = () => {
  let timeAfter30Mins = new Date();
  timeAfter30Mins = new Date(timeAfter30Mins.setMinutes(timeAfter30Mins.getMinutes() + 30));
};

这样称呼它:

getTimeAfter30Mins();

事情就是这么简单;

let initial_date = new Date;
let added30Min = new Date(initial_date.getTime() + (30*60*1000));

最简单的解决方法是认识到在javascript中日期只是数字。它开始于1969年12月31日星期三18:00:00 GMT-0600 (CST)。每1代表一毫秒。您可以通过获取值并使用该值实例化一个新日期来增加或减少毫秒数。用这种思维,你很容易就能搞定。

const minutesToAdjust = 10;
const millisecondsPerMinute = 60000;
const originalDate = new Date('11/20/2017 10:00 AM');
const modifiedDate1 = new Date(originalDate.valueOf() - (minutesToAdjust * millisecondsPerMinute));
const modifiedDate2 = new Date(originalDate.valueOf() + (minutesToAdjust * millisecondsPerMinute));

console.log(originalDate); // Mon Nov 20 2017 10:00:00 GMT-0600 (CST)
console.log(modifiedDate1); // Mon Nov 20 2017 09:50:00 GMT-0600 (CST)
console.log(modifiedDate2); // Mon Nov 20 2017 10:10:00 GMT-0600 (CST)

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>

var d1 = new Date (),
    d2 = new Date ( d1 );
d2.setMinutes ( d1.getMinutes() + 30 );
alert ( d2 );