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
我认为解决方案不是100%正确的，我会使用天花板而不是地板，圆形将工作，但这不是正确的操作。
function dateDiff(str1, str2){
var diff = Date.parse(str2) - Date.parse(str1);
return isNaN(diff) ? NaN : {
diff: diff,
ms: Math.ceil(diff % 1000),
s: Math.ceil(diff / 1000 % 60),
m: Math.ceil(diff / 60000 % 60),
h: Math.ceil(diff / 3600000 % 24),
d: Math.ceil(diff / 86400000)
};
}
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;
}
}
1970-01-01之前和2038-01-19之后的贡献
function DateDiff(aDate1, aDate2) {
let dDay = 0;
this.isBissexto = (aYear) => {
return (aYear % 4 == 0 && aYear % 100 != 0) || (aYear % 400 == 0);
};
this.getDayOfYear = (aDate) => {
let count = 0;
for (let m = 0; m < aDate.getUTCMonth(); m++) {
count += m == 1 ? this.isBissexto(aDate.getUTCFullYear()) ? 29 : 28 : /(3|5|8|10)/.test(m) ? 30 : 31;
}
count += aDate.getUTCDate();
return count;
};
this.toDays = () => {
return dDay;
};
(() => {
let startDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate1.toISOString()) : new Date(aDate2.toISOString());
let endDate = aDate1.getTime() <= aDate2.getTime() ? new Date(aDate2.toISOString()) : new Date(aDate1.toISOString());
while (startDate.getUTCFullYear() != endDate.getUTCFullYear()) {
dDay += (this.isBissexto(startDate.getFullYear())? 366 : 365) - this.getDayOfYear(startDate) + 1;
startDate = new Date(startDate.getUTCFullYear()+1, 0, 1);
}
dDay += this.getDayOfYear(endDate) - this.getDayOfYear(startDate);
})();
}
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
我建议使用moment.js库(http://momentjs.com/docs/#/displaying/difference/)。它正确地处理夏令时，通常是很好的工作。
例子:
var start = moment("2013-11-03");
var end = moment("2013-11-04");
end.diff(start, "days")
1
我认为解决方案不是100%正确的，我会使用天花板而不是地板，圆形将工作，但这不是正确的操作。
function dateDiff(str1, str2){
var diff = Date.parse(str2) - Date.parse(str1);
return isNaN(diff) ? NaN : {
diff: diff,
ms: Math.ceil(diff % 1000),
s: Math.ceil(diff / 1000 % 60),
m: Math.ceil(diff / 60000 % 60),
h: Math.ceil(diff / 3600000 % 24),
d: Math.ceil(diff / 86400000)
};
}
