问题描述
您好我正在使用jquery-ui datepicker选择日期并找出2个日期之间的差异。
Hi i am using jquery-ui datepicker to select date and date.js to find difference between 2 dates.
现在问题是我想从计算中排除周末天数(星期六和星期日)。我该怎么做?
Right now the problem is I want to exclude weekend days from calculation (saturday and sunday). How should i do that?
例如,用户选择开始日期(13/8/2010)和结束日期(16/8/2010)。由于2010年8月14日和15/8/2010是在工作日,而不是总共4天,我希望它只有2天。
For example the user select start date (13/8/2010) and end date (16/8/2010). Since 14/8/2010 and 15/8/2010 is in week days, instead of 4 days total, i want it to be only 2 days.
这是代码即时使用:
<script type="text/javascript">
$("#startdate, #enddate").change(function() {
var d1 = $("#startdate").val();
var d2 = $("#enddate").val();
var minutes = 1000*60;
var hours = minutes*60;
var day = hours*24;
var startdate1 = getDateFromFormat(d1, "d-m-y");
var enddate1 = getDateFromFormat(d2, "d-m-y");
var days = 1 + Math.round((enddate1 - startdate1)/day);
if(days>0)
{ $("#noofdays").val(days);}
else
{ $("#noofdays").val(0);}
});
</script>
推荐答案
也许其他人可以帮助您将此功能转换为JQuery的框架...
Maybe someone else can help you converting this function into JQuery's framework...
在原始javascript中,我将使用:
In raw javascript i will use:
<script>
function calcBusinessDays(dDate1, dDate2) { // input given as Date objects
var iWeeks, iDateDiff, iAdjust = 0;
if (dDate2 < dDate1) return -1; // error code if dates transposed
var iWeekday1 = dDate1.getDay(); // day of week
var iWeekday2 = dDate2.getDay();
iWeekday1 = (iWeekday1 == 0) ? 7 : iWeekday1; // change Sunday from 0 to 7
iWeekday2 = (iWeekday2 == 0) ? 7 : iWeekday2;
if ((iWeekday1 > 5) && (iWeekday2 > 5)) iAdjust = 1; // adjustment if both days on weekend
iWeekday1 = (iWeekday1 > 5) ? 5 : iWeekday1; // only count weekdays
iWeekday2 = (iWeekday2 > 5) ? 5 : iWeekday2;
// calculate differnece in weeks (1000mS * 60sec * 60min * 24hrs * 7 days = 604800000)
iWeeks = Math.floor((dDate2.getTime() - dDate1.getTime()) / 604800000)
if (iWeekday1 <= iWeekday2) {
iDateDiff = (iWeeks * 5) + (iWeekday2 - iWeekday1)
} else {
iDateDiff = ((iWeeks + 1) * 5) - (iWeekday1 - iWeekday2)
}
iDateDiff -= iAdjust // take into account both days on weekend
return (iDateDiff + 1); // add 1 because dates are inclusive
}
</script>
发现
并调用该函数,例如:
<script>
alert(calcBusinessDays(new Date("August 11, 2010 11:13:00"),new Date("August 16, 2010 11:13:00")));
</script>
## EDITED ##
如果你只想使用它的格式:
## EDITED ##
If you want to use it with your that format just:
你的代码如下:
<script type="text/javascript">
$("#startdate, #enddate").change(function() {
var d1 = $("#startdate").val();
var d2 = $("#enddate").val();
var minutes = 1000*60;
var hours = minutes*60;
var day = hours*24;
var startdate1 = getDateFromFormat(d1, "d-m-y");
var enddate1 = getDateFromFormat(d2, "d-m-y");
var newstartdate=new Date();
newstartdate.setFullYear(startdate1.getYear(),startdate1.getMonth(),startdate1.getDay());
var newenddate=new Date();
newenddate.setFullYear(enddate1.getYear(),enddate1.getMonth(),enddate1.getDay());
var days = calcBusinessDays(newstartdate,newenddate);
if(days>0)
{ $("#noofdays").val(days);}
else
{ $("#noofdays").val(0);}
});
</script>
这篇关于查找两个日期之间的差异(周末除外)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!