这里是一个函数,这将帮助你.. 它的用法相同的SQL ..
function datediff(fromDate,toDate,interval) {
var second=1000, minute=second*60, hour=minute*60, day=hour*24, week=day*7;
fromDate = new Date(fromDate);
toDate = new Date(toDate);
var timediff = toDate - fromDate;
if (isNaN(timediff)) return NaN;
switch (interval) {
case "years": return toDate.getFullYear() - fromDate.getFullYear();
case "months": return (
(toDate.getFullYear() * 12 + toDate.getMonth())
-
(fromDate.getFullYear() * 12 + fromDate.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;
}
}
向下滚动更新的提琴在这里看到“整洁”版本; http://jsfiddle.net/uUqrT/5/
$("#calc").click(function()
{
addDaysRow($('#from').val(), $('#to').val());
});
function addDaysRow(fromDate, toDate) {
var rangeAstart = new Date('01/01/2012');
var rangeAend = new Date('06/30/2012');
var rangeBstart = new Date('07/01/2012');
var rangeBend = new Date('12/31/2012');
var rangeCstart = new Date('01/01/2013');
var rangeCend = new Date('06/30/2013');
var rangeDstart = new Date('07/01/2013');
var rangeDend = new Date('12/31/2013');
var diff = datediff(fromDate, toDate, "days");
$('#record > tbody:last').append('<tr><td>' + diff + '</td><td>' + diff + '</td><td>' +diff + '</td><td>' + diff + '</td><td>' + diff + '</td><td>' + diff + '</td><td>' + diff + '</td></tr>');
}
这应该在正确的轨道上开始你 - 不知道你想在这里与范围内做什么。
DATEDIFF('10/10/2010' ,'20/12/2010' , '天' ); –
这个函数计算两个日期之间的天数,感谢这个。但如何计算谷底所有区间的天数?例如:返回范围A中的天数,比范围B中的天数,范围内的C等...我有50个范围或多或少,我必须使50 var?有没有办法使用循环?谢谢 – fasenderos
是的,你在做什么?你可以在JSFiddle中做一个模型,并告诉我你需要什么功能以及在哪里?然后我会填写代码。 @fasenderos –