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繁星coding
var oneDay = 24*60*60*1000; // hours*minutes*seconds*millisecondsvar firstDate = new Date(2008,01,12);var secondDate = new Date(2008,01,22);
var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
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素胚勾勒不出你
下面是一个这样做的函数：function days_between(date1, date2) {
// The number of milliseconds in one day
var ONE_DAY = 1000 * 60 * 60 * 24;
// Convert both dates to milliseconds
var date1_ms = date1.getTime();
var date2_ms = date2.getTime();
// Calculate the difference in milliseconds
var difference_ms = Math.abs(date1_ms - date2_ms);
// Convert back to days and return
return Math.round(difference_ms/ONE_DAY);}
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动漫人物
我用的是这个。如果您只是减去日期，它将无法跨夏令节约时间边界(如4月1日至4月30日或10月1日至10月31日)。这减少了所有的时间，以确保您有一天，并消除任何DST问题使用UTC。var nDays = ( Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) -
Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;作为一项职能：function DaysBetween(StartDate, EndDate) {
// The number of milliseconds in all UTC days (no DST)
const oneDay = 1000 * 60 * 60 * 24;
// A day in UTC always lasts 24 hours (unlike in other time formats)
const start = Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate());
const end = Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate());
// so it's safe to divide by 24 hours
return (start - end) / oneDay;}