Find day difference between two dates (excluding weekend days)
Maybe someone else can help you converting this function into JQuery's framework...
I found this function here.
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) { //Equal to makes it reduce 5 days
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
}
var date1 = new Date("August 11, 2010 11:13:00");
var date2 = new Date("August 16, 2010 11:13:00");
alert(calcBusinessDays(date1, date2));## EDITED ##
If you want to use it with your that format just:
Your code will look like:
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) { //Equal to makes it reduce 5 days
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
}
$("#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 = new Date(d1);
var enddate1 = new Date(d2);
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 src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
<label>Start Date
<input type="date" id="startdate" value="2019-03-03"/>
</label>
<label>End Date
<input type="date" id="enddate" value="2019-03-06"/>
</label>
<label>N. of days
<output id="noofdays"/>
</label>To do this, you should NOT search all days between these dates !
It's not complicated, look some evident assumptions:
All full-week has 7-days.
Which 2 are weekend-days.
And which 5 are business-day.
Evident conclusions:
Look all days is loss of time.
Check what day is weekend to all week is loss of time.
Without tedious explanation.. let me show the code:
function getBusinessDateCount (startDate, endDate) {
var elapsed, daysBeforeFirstSaturday, daysAfterLastSunday;
var ifThen = function (a, b, c) {
return a == b ? c : a;
};
elapsed = endDate - startDate;
elapsed /= 86400000;
daysBeforeFirstSunday = (7 - startDate.getDay()) % 7;
daysAfterLastSunday = endDate.getDay();
elapsed -= (daysBeforeFirstSunday + daysAfterLastSunday);
elapsed = (elapsed / 7) * 5;
elapsed += ifThen(daysBeforeFirstSunday - 1, -1, 0) + ifThen(daysAfterLastSunday, 6, 5);
return Math.ceil(elapsed);
}
function calc() {
let start = document.querySelector('#startDate').value,
end = document.querySelector('#endDate').value,
result = getBusinessDateCount(new Date(start), new Date(end));
document.querySelector('#result').value = result;
}Start date: <input type="date" id="startDate" value="2020-01-04"><br>
End date: <input type="date" id="endDate" value="2020-01-06"><br>
<input type="button" onclick="calc()" value="Get business days"><br>
Business days: <input id="result" readonly>You can test it yourself with any dates.
I just want to notice that this code ONLY consumed 0.43 sec between dates from 2000 to 2015... It is much more fast than some other codes.
Hope it helps...
Nice coding !!
This is how I would do it
function getDays(d1, d2) {
var one_day=1000*60*60*24;
var d1_days = parseInt(d1.getTime()/one_day) - 1;
var d2_days = parseInt(d2.getTime()/one_day);
var days = (d2_days - d1_days);
var weeks = (d2_days - d1_days) / 7;
var day1 = d1.getDay();
var day2 = d2.getDay();
if (day1 == 0) {
days--;
} else if (day1 == 6) {
days-=2;
}
if (day2 == 0) {
days-=2;
} else if (day2 == 6) {
days--;
}
days -= parseInt(weeks) * 2;
alert(days);
}
getDays(new Date("June 8, 2004"),new Date("February 6, 2010"));
EDIT
To clarify my comment to @keenebec...
That solution will work for small date differences quite nicely and is easy to understand. But take something as "short" as a 6 year span and you can see a remarkable difference in speed.
http://jsfiddle.net/aSvxv/
I included all 3 answers and the original answer is indeed the fastest, but not by much and the trade off for a few microseconds of execution is somewhat trivial to me in favor of readability.
Date.prototype.addDays = function(days) {
var date = new Date(this.valueOf())
date.setDate(date.getDate() + days);
return date;
}
function getBusinessDatesCount(startDate, endDate) {
var count = 0;
var curDate = startDate;
while (curDate <= endDate) {
var dayOfWeek = curDate.getDay();
var isWeekend = (dayOfWeek == 6) || (dayOfWeek == 0);
if(!isWeekend)
count++;
curDate = curDate.addDays(1);
}
return count;
}
//Usage
var startDate = new Date('7/16/2015');
var endDate = new Date('7/20/2015');
var numOfDays = getBusinessDatesCount(startDate,endDate);
jQuery('div#result').text(numOfDays);<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="result"/>