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?
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
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。
其他回答
一行代码和小代码
const diff=(e,t)=>Math.floor((new Date(e).getTime()-new Date(t).getTime())/1000*60*60*24);
// or
const diff=(e,t)=>Math.floor((new Date(e)-new Date(t))/864e5);
// or
const diff=(a,b)=>(new Date(a)-new Date(b))/864e5|0;
// use
diff('1/1/2001', '1/1/2000')
为打印稿
const diff = (from: string, to: string) => Math.floor((new Date(from).getTime() - new Date(to).getTime()) / 86400000);
从DatePicker小部件使用formatDate怎么样?您可以使用它来转换时间戳格式的日期(从01/01/1970开始的毫秒),然后做一个简单的减法。
function validateDate() {
// get dates from input fields
var startDate = $("#startDate").val();
var endDate = $("#endDate").val();
var sdate = startDate.split("-");
var edate = endDate.split("-");
var diffd = (edate[2] - sdate[2]) + 1;
var leap = [ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
var nonleap = [ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
if (sdate[0] > edate[0]) {
alert("Please enter End Date Year greater than Start Date Year");
document.getElementById("endDate").value = "";
diffd = "";
} else if (sdate[1] > edate[1]) {
alert("Please enter End Date month greater than Start Date month");
document.getElementById("endDate").value = "";
diffd = "";
} else if (sdate[2] > edate[2]) {
alert("Please enter End Date greater than Start Date");
document.getElementById("endDate").value = "";
diffd = "";
} else {
if (sdate[0] / 4 == 0) {
while (sdate[1] < edate[1]) {
diffd = diffd + leap[sdate[1]++];
}
} else {
while (sdate[1] < edate[1]) {
diffd = diffd + nonleap[sdate[1]++];
}
}
document.getElementById("numberOfDays").value = diffd;
}
}
试试这个
let today = new Date(). toisostring()。片(0,10) const startDate = '2021-04-15'; const endDate = today; const diffInMs = new Date(endDate) - new Date(startDate) const diffInDays = diffInMs / (1000 * 60 * 60 * 24); alert(diffInDays);
我使用下面的代码来试验新闻帖子的发布日期功能。我根据发布日期和当前日期计算分钟、小时、天或年。
var startDate= new Date("Mon Jan 01 2007 11:00:00");
var endDate =new Date("Tue Jan 02 2007 12:50:00");
var timeStart = startDate.getTime();
var timeEnd = endDate.getTime();
var yearStart = startDate.getFullYear();
var yearEnd = endDate.getFullYear();
if(yearStart == yearEnd)
{
var hourDiff = timeEnd - timeStart;
var secDiff = hourDiff / 1000;
var minDiff = hourDiff / 60 / 1000;
var hDiff = hourDiff / 3600 / 1000;
var myObj = {};
myObj.hours = Math.floor(hDiff);
myObj.minutes = minDiff
if(myObj.hours >= 24)
{
console.log(Math.floor(myObj.hours/24) + "day(s) ago")
}
else if(myObj.hours>0)
{
console.log(myObj.hours +"hour(s) ago")
}
else
{
console.log(Math.abs(myObj.minutes) +"minute(s) ago")
}
}
else
{
var yearDiff = yearEnd - yearStart;
console.log( yearDiff +" year(s) ago");
}