我有两个任意日期,比如2012年4月1日和2013年1月15日。我想计算这两个日期之间的星期日,星期一,星期二,星期三,星期四,星期五和星期六的数量。
有没有一种可靠的快速方法来做到这一点而不会削弱用户的CPU /浏览器?
由于
更新
这样做的前提是,我们为一周中的任何一天定义了平均事件数。我们需要计算在一段时间内发生的事件数量,即使是部分事件(例如星期日的1/2天将是添加到总数的事件数量的一半)
答案 0 :(得分:2)
好的,这是一个可能未经测试的解决方案
date1 = new Date("2012-02-10");
date2 = new Date("2012-03-10");
daysInBetween = (date2.getTime() - date1.getTime())/1000/3600/24;
dayOfTheWeek1 = date1.getDay();
weeks = parseInt(daysInBetween/7, 10);
extraDays = daysInBetween%7;
您有weeks + 1
天dayOfWeek1 ... dayOfWeek1 + (6 - extraDays)
您有weeks + 1 + extraDays
天dayOfWeek1 + (6 - extraDays) ... dayOfWeek1 + 6
请注意,如果dayOfWeek1 === 6
,那么我假设dayOfWeek1 + 1 === 0
。
编辑:
更多代码:
var days = {};
var dayOfTheWeekEnd = dayOfTheWeek1 + 6 - extraDays; // no imagination for names...
if (dayOfTheWeekEnd < 6) {
if (0 >= dayOfTheWeek1 && 0 <= dayOfTheWeekEnd) {
days.sunday = weeks + 1;
} else {
days.sunday = weeks + 1 + extraDays;
}
// etc for the other days, a for loog with an i instead of the 0 would be better.
} else {
// I have to go the school! I'll edit it later.
// The idea is that you have to take dayOfTheWeekEnd back to the 0-6 range
// and check if its after dayOfWeek1 or before dayOfTheWeekEnd, then days.sunday=weeks+1.
}
答案 1 :(得分:0)
通过思考......没有测试过,但我希望它有所帮助。
function(date1, date2) {
// date1 is assumed to be 12:00:00 am and date2 is actually the date after target at
// 12:00:00 am to make the dates inclusive so from the example date1 = 2012/4/1 and
// date2 = 2013/1/16
var timeBetween = date2.getTime() - date1.getTime(); //milliseconds between
var weeks = timeBetween/1000/60/60/24/7;
var sun = weeks;
var mon = weeks;
var tue = weeks;
var wed = weeks;
var thu = weeks;
var fri = weeks;
var sat = weeks;
date1.setTime(date1.getTime() + weeks*7*24*60*60*1000); //set the first date to x weeks later
//if this iterates more than 7 times then something is wrong.
while(date1.getTime() < date2.getTime()){
switch(date1.getDay()){
case 0:
sun++;
break;
case 1:
mon++;
break;
...
}
date1.setTime(date1.getTime() + 1000 * 60 * 60 * 24); //Move date 1 to the next date.
}
}