例如，在输入框中给定两个日期:
<input id="first" value="1/1/2000"/>
<input id="second" value="1/1/2001"/>
<script>
alert(datediff("day", first, second)); // what goes here?
</script>
如何在JavaScript中获得两个日期之间的天数?
当前回答
// JavaScript / NodeJs answer
let startDate = new Date("2022-09-19");
let endDate = new Date("2022-09-26");
let difference = startDate.getTime() - endDate.getTime();
console.log(difference);
let TotalDiffDays = Math.ceil(difference / (1000 * 3600 * 24));
console.log(TotalDiffDays + " days :) ");
其他回答
使用Moment.js
Var future = moment('05/02/2015'); Var start = moment('04/23/2015'); Var d =未来。diff(开始,“天”);/ / 9 console.log (d); < script src = " https://cdnjs.cloudflare.com/ajax/libs/moment.js/2.17.1/moment-with-locales.min.js " > < /脚本>
最好还是取消夏令时吧，马斯。装天花板,数学。楼层等，使用UTC时间:
var firstDate = Date.UTC(2015,01,2);
var secondDate = Date.UTC(2015,04,22);
var diff = Math.abs((firstDate.valueOf()
- secondDate.valueOf())/(24*60*60*1000));
这个例子给出了109天的差异。24*60*60*1000是一天，单位是毫秒。
在撰写本文时，其他答案中只有一个正确处理DST(夏令时)转换。以下是位于加州的一个系统的结果:
1/1/2013- 3/10/2013- 11/3/2013-
User Formula 2/1/2013 3/11/2013 11/4/2013 Result
--------- --------------------------- -------- --------- --------- ---------
Miles (d2 - d1) / N 31 0.9583333 1.0416666 Incorrect
some Math.floor((d2 - d1) / N) 31 0 1 Incorrect
fuentesjr Math.round((d2 - d1) / N) 31 1 1 Correct
toloco Math.ceiling((d2 - d1) / N) 31 1 2 Incorrect
N = 86400000
虽然数学。round返回正确的结果，我认为它有点笨拙。相反，当DST开始或结束时，通过显式计算UTC偏移量的变化，我们可以使用精确的算术:
function treatAsUTC(date) {
var result = new Date(date);
result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
return result;
}
function daysBetween(startDate, endDate) {
var millisecondsPerDay = 24 * 60 * 60 * 1000;
return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}
alert(daysBetween($('#first').val(), $('#second').val()));
解释
JavaScript的日期计算很棘手，因为date对象内部存储的时间是UTC，而不是本地时间。例如，3/10/2013太平洋标准时间12:00 AM (UTC-08:00)存储为3/10/2013上午8:00 UTC, 3/11/2013太平洋夏令时12:00 AM (UTC-07:00)存储为3/11/2013上午7:00 UTC。在这一天，从午夜到午夜，当地时间在UTC只有23小时!
虽然本地时间中的一天可以大于或小于24小时，但国际标准时间中的一天总是24小时上面所示的daysBetween方法利用了这一事实，它首先调用treatAsUTC将本地时间调整为午夜UTC，然后再进行减法和除法。
1. JavaScript忽略闰秒。
function formatDate(seconds, dictionary) {
var foo = new Date;
var unixtime_ms = foo.getTime();
var unixtime = parseInt(unixtime_ms / 1000);
var diff = unixtime - seconds;
var display_date;
if (diff <= 0) {
display_date = dictionary.now;
} else if (diff < 60) {
if (diff == 1) {
display_date = diff + ' ' + dictionary.second;
} else {
display_date = diff + ' ' + dictionary.seconds;
}
} else if (diff < 3540) {
diff = Math.round(diff / 60);
if (diff == 1) {
display_date = diff + ' ' + dictionary.minute;
} else {
display_date = diff + ' ' + dictionary.minutes;
}
} else if (diff < 82800) {
diff = Math.round(diff / 3600);
if (diff == 1) {
display_date = diff + ' ' + dictionary.hour;
} else {
display_date = diff + ' ' + dictionary.hours;
}
} else {
diff = Math.round(diff / 86400);
if (diff == 1) {
display_date = diff + ' ' + dictionary.day;
} else {
display_date = diff + ' ' + dictionary.days;
}
}
return display_date;
}
使用以下公式可以计算出两个不同TZs的休止日期之间的完整证明天数差:
var start = new Date('10/3/2015'); var end = new Date('11/2/2015'); Var天数= (end - start) / 1000 / 60 / 60 / 24; console.log(天); //实际上是30;但由于夏令时，将显示31.0xxx //你需要抵消如下 days = days - (end.getTimezoneOffset() - start.getTimezoneOffset()) / (60 * 24); console.log(天);
