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
区分两个日期的最简单方法是:
var diff = Math.floor((Date.parse(str2) - Date.parse(str1)) / 86400000);
您将得到不同的天数(如果其中一个或两个都无法解析,则为NaN)。解析日期给出了以毫秒为单位的结果,要按天得到它,你必须除以24 * 60 * 60 * 1000
如果你想用天、小时、分钟、秒和毫秒来划分:
function dateDiff( str1, str2 ) {
var diff = Date.parse( str2 ) - Date.parse( str1 );
return isNaN( diff ) ? NaN : {
diff : diff,
ms : Math.floor( diff % 1000 ),
s : Math.floor( diff / 1000 % 60 ),
m : Math.floor( diff / 60000 % 60 ),
h : Math.floor( diff / 3600000 % 24 ),
d : Math.floor( diff / 86400000 )
};
}
以下是我对James版本的重构版本:
function mydiff(date1,date2,interval) {
var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
date1 = new Date(date1);
date2 = new Date(date2);
var timediff = date2 - date1;
if (isNaN(timediff)) return NaN;
switch (interval) {
case "years": return date2.getFullYear() - date1.getFullYear();
case "months": return (
( date2.getFullYear() * 12 + date2.getMonth() )
-
( date1.getFullYear() * 12 + date1.getMonth() )
);
case "weeks" : return Math.floor(timediff / week);
case "days" : return Math.floor(timediff / day);
case "hours" : return Math.floor(timediff / hour);
case "minutes": return Math.floor(timediff / minute);
case "seconds": return Math.floor(timediff / second);
default: return undefined;
}
}
其他回答
区分两个日期的最简单方法是:
var diff = Math.floor((Date.parse(str2) - Date.parse(str1)) / 86400000);
您将得到不同的天数(如果其中一个或两个都无法解析,则为NaN)。解析日期给出了以毫秒为单位的结果,要按天得到它,你必须除以24 * 60 * 60 * 1000
如果你想用天、小时、分钟、秒和毫秒来划分:
function dateDiff( str1, str2 ) {
var diff = Date.parse( str2 ) - Date.parse( str1 );
return isNaN( diff ) ? NaN : {
diff : diff,
ms : Math.floor( diff % 1000 ),
s : Math.floor( diff / 1000 % 60 ),
m : Math.floor( diff / 60000 % 60 ),
h : Math.floor( diff / 3600000 % 24 ),
d : Math.floor( diff / 86400000 )
};
}
以下是我对James版本的重构版本:
function mydiff(date1,date2,interval) {
var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
date1 = new Date(date1);
date2 = new Date(date2);
var timediff = date2 - date1;
if (isNaN(timediff)) return NaN;
switch (interval) {
case "years": return date2.getFullYear() - date1.getFullYear();
case "months": return (
( date2.getFullYear() * 12 + date2.getMonth() )
-
( date1.getFullYear() * 12 + date1.getMonth() )
);
case "weeks" : return Math.floor(timediff / week);
case "days" : return Math.floor(timediff / day);
case "hours" : return Math.floor(timediff / hour);
case "minutes": return Math.floor(timediff / minute);
case "seconds": return Math.floor(timediff / second);
default: return undefined;
}
}
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。
简单、容易、复杂。此函数将每1秒调用一次以更新时间。
const year = (new Date().getFullYear());
const bdayDate = new Date("04,11,2019").getTime(); //mmddyyyy
// countdown
let timer = setInterval(function () {
// get today's date
const today = new Date().getTime();
// get the difference
const diff = bdayDate - today;
// math
let days = Math.floor(diff / (1000 * 60 * 60 * 24));
let hours = Math.floor((diff % (1000 * 60 * 60 * 24)) / (1000 * 60 * 60));
let minutes = Math.floor((diff % (1000 * 60 * 60)) / (1000 * 60));
let seconds = Math.floor((diff % (1000 * 60)) / 1000);
}, 1000);
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);
如果你想有一个DateArray日期试试这个:
<script>
function getDates(startDate, stopDate) {
var dateArray = new Array();
var currentDate = moment(startDate);
dateArray.push( moment(currentDate).format('L'));
var stopDate = moment(stopDate);
while (dateArray[dateArray.length -1] != stopDate._i) {
dateArray.push( moment(currentDate).format('L'));
currentDate = moment(currentDate).add(1, 'days');
}
return dateArray;
}
</script>
调试片段