如何计算两个date之间的天数?
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我正在计算“从”到“到”date之间的天数。 例如,如果起始date是13/04/2010,并且截止date是15/04/2010,则结果应该是
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如何使用JavaScript获得结果?
谷歌search“两个date之间的datejavascript”产生这个伟大的片段 (实际上所有的顶部结果都与您的问题相关):
var oneDay = 24*60*60*1000; // hours*minutes*seconds*milliseconds var firstDate = new Date(2008,01,12); var secondDate = new Date(2008,01,22); var diffDays = Math.round(Math.abs((firstDate.getTime() - secondDate.getTime())/(oneDay)));
这是一个这样的function :
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) }
调整为允许夏时制差异。 尝试这个:
function daysBetween(date1, date2) { // adjust diff for for daylight savings var hoursToAdjust = Math.abs(date1.getTimezoneOffset() /60) - Math.abs(date2.getTimezoneOffset() /60); // apply the tz offset date2.addHours(hoursToAdjust); // 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) } // you'll want this addHours function too Date.prototype.addHours= function(h){ this.setHours(this.getHours()+h); return this; }
这是我使用的。 如果你只是减去date,它将不会跨越夏令时边界(例如4月1日至4月30日或10月1日至10月31日)。 这会使所有的小时数下降,以确保您获得一天,并通过使用UTC消除任何DST问题。
var nDays = ( Date.UTC(EndDate.getFullYear(), EndDate.getMonth(), EndDate.getDate()) - Date.UTC(StartDate.getFullYear(), StartDate.getMonth(), StartDate.getDate())) / 86400000;
这是我的实现:
function daysBetween(one, another) { return Math.round(Math.abs((+one) - (+another))/8.64e7); }
+<date>
将types强制转换为整数表示forms,其作用与<date>.getTime()
和8.64e7
是一天中的毫秒数相同。
我为另一个post写了这个解决scheme,如何计算两个date之间的差异,所以我分享了我所准备的:
// Here are the two dates to compare var date1 = '2011-12-24'; var date2 = '2012-01-01'; // First we split the values to arrays date1[0] is the year, [1] the month and [2] the day date1 = date1.split('-'); date2 = date2.split('-'); // Now we convert the array to a Date object, which has several helpful methods date1 = new Date(date1[0], date1[1], date1[2]); date2 = new Date(date2[0], date2[1], date2[2]); // We use the getTime() method and get the unixtime (in milliseconds, but we want seconds, therefore we divide it through 1000) date1_unixtime = parseInt(date1.getTime() / 1000); date2_unixtime = parseInt(date2.getTime() / 1000); // This is the calculated difference in seconds var timeDifference = date2_unixtime - date1_unixtime; // in Hours var timeDifferenceInHours = timeDifference / 60 / 60; // and finaly, in days :) var timeDifferenceInDays = timeDifferenceInHours / 24; alert(timeDifferenceInDays);
你可以跳过代码中的一些步骤,我已经写了它,以便于理解。
你会在这里find一个运行的例子: http : //jsfiddle.net/matKX/
从我的小date差异计算器:
var startDate = new Date(2000, 1-1, 1); // 2000-01-01 var endDate = new Date(); // Today // Calculate the difference of two dates in total days function diffDays(d1, d2) { var ndays; var tv1 = d1.valueOf(); // msec since 1970 var tv2 = d2.valueOf(); ndays = (tv2 - tv1) / 1000 / 86400; ndays = Math.round(ndays - 0.5); return ndays; }
所以你会打电话给:
var nDays = diffDays(startDate, endDate);
(完整资料来自http://david.tribble.com/src/javascript/jstimespan.html )
附录
代码可以通过改变这些行来改进:
var tv1 = d1.getTime(); // msec since 1970 var tv2 = d2.getTime();