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
我也有同样的问题，但如果你在SQL查询上做的话会更好:
DateDiff(DAY, StartValue,GETDATE()) AS CountDays
查询将自动生成一个列CountDays
其他回答
Date.prototype.days =函数(到){ 返回Math.abs(Math.floor(to.getTime() / (3600 * 24 * 1000)) - Math.floor(this.getTime() / (3600 * 24 * 1000))) ｝ console.log(新日期(“2014/05/20”)。天(新日期(2014/05/23)));// 3天 console.log(新日期(“2014/05/23”)。天(新日期(2014/05/20)));// 3天
我也有同样的问题，但如果你在SQL查询上做的话会更好:
DateDiff(DAY, StartValue,GETDATE()) AS CountDays
查询将自动生成一个列CountDays
function validateDate() {
// get dates from input fields
var startDate = $("#startDate").val();
var endDate = $("#endDate").val();
var sdate = startDate.split("-");
var edate = endDate.split("-");
var diffd = (edate[2] - sdate[2]) + 1;
var leap = [ 0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
var nonleap = [ 0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31 ];
if (sdate[0] > edate[0]) {
alert("Please enter End Date Year greater than Start Date Year");
document.getElementById("endDate").value = "";
diffd = "";
} else if (sdate[1] > edate[1]) {
alert("Please enter End Date month greater than Start Date month");
document.getElementById("endDate").value = "";
diffd = "";
} else if (sdate[2] > edate[2]) {
alert("Please enter End Date greater than Start Date");
document.getElementById("endDate").value = "";
diffd = "";
} else {
if (sdate[0] / 4 == 0) {
while (sdate[1] < edate[1]) {
diffd = diffd + leap[sdate[1]++];
}
} else {
while (sdate[1] < edate[1]) {
diffd = diffd + nonleap[sdate[1]++];
}
}
document.getElementById("numberOfDays").value = diffd;
}
}
在撰写本文时，其他答案中只有一个正确处理DST(夏令时)转换。以下是位于加州的一个系统的结果:
1/1/2013- 3/10/2013- 11/3/2013-
User Formula 2/1/2013 3/11/2013 11/4/2013 Result
--------- --------------------------- -------- --------- --------- ---------
Miles (d2 - d1) / N 31 0.9583333 1.0416666 Incorrect
some Math.floor((d2 - d1) / N) 31 0 1 Incorrect
fuentesjr Math.round((d2 - d1) / N) 31 1 1 Correct
toloco Math.ceiling((d2 - d1) / N) 31 1 2 Incorrect
N = 86400000
虽然数学。round返回正确的结果，我认为它有点笨拙。相反，当DST开始或结束时，通过显式计算UTC偏移量的变化，我们可以使用精确的算术:
function treatAsUTC(date) {
var result = new Date(date);
result.setMinutes(result.getMinutes() - result.getTimezoneOffset());
return result;
}
function daysBetween(startDate, endDate) {
var millisecondsPerDay = 24 * 60 * 60 * 1000;
return (treatAsUTC(endDate) - treatAsUTC(startDate)) / millisecondsPerDay;
}
alert(daysBetween($('#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。
我从其他答案中得到一些灵感，使输入具有自动卫生。我希望这是对其他答案的改进。
我还推荐使用<input type="date">字段，这将有助于验证用户输入。
//use best practices by labeling your constants. let MS_PER_SEC = 1000 , SEC_PER_HR = 60 * 60 , HR_PER_DAY = 24 , MS_PER_DAY = MS_PER_SEC * SEC_PER_HR * HR_PER_DAY ; //let's assume we get Date objects as arguments, otherwise return 0. function dateDiffInDays(date1Time, date2Time) { if (!date1Time || !date2Time) return 0; return Math.round((date2Time - date1Time) / MS_PER_DAY); } function getUTCTime(dateStr) { const date = new Date(dateStr); // If use 'Date.getTime()' it doesn't compute the right amount of days // if there is a 'day saving time' change between dates return Date.UTC(date.getFullYear(), date.getMonth(), date.getDate()); } function calcInputs() { let date1 = document.getElementById("date1") , date2 = document.getElementById("date2") , resultSpan = document.getElementById("result") ; if (date1.value && date2.value && resultSpan) { //remove non-date characters console.log(getUTCTime(date1.value)); let date1Time = getUTCTime(date1.value) , date2Time = getUTCTime(date2.value) , result = dateDiffInDays(date1Time, date2Time) ; resultSpan.innerHTML = result + " days"; } } window.onload = function() { calcInputs(); }; //some code examples console.log(dateDiffInDays(new Date("1/15/2019"), new Date("1/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2019"), new Date("2/30/2019"))); console.log(dateDiffInDays(new Date("1/15/2000"), new Date("1/15/2019"))); <input type="date" id="date1" value="2000-01-01" onchange="calcInputs();" /> <input type="date" id="date2" value="2022-01-01" onchange="calcInputs();"/> Result: <span id="result"></span>
