I have two sets of codes that work. Needed help combining them into one.
This code gets me the difference between two dates. works perfectly:
functio
When determining the number of days between two dates, there are lots of decisions to be made about what is a day. For example, the period 1 Feb to 2 Feb is generally one day, so 1 Feb to 1 Feb is zero days.
When adding the complexity of counting only business days, things get a lot tougher. E.g. Monday 2 Feb 2015 to Friday 6 February is 4 elapsed days (Monday to Tuesday is 1, Monday to Wednesday is 2, etc.), however the expression "Monday to Friday" is generally viewed as 5 business days and the duration Mon 2 Feb to Sat 7 Feb should also be 4 business days, but Sunday to Saturday should be 5.
So here's my algorithm:
The stepping part at the end can probably be replaced by some other algorithm, but it will never loop for more than 6 days so it's a simple and reasonably efficient solution to the issue of uneven weeks.
Some consequences of the above:
Here's the code:
// Expects start date to be before end date
// start and end are Date objects
function dateDifference(start, end) {
// Copy date objects so don't modify originals
var s = new Date(+start);
var e = new Date(+end);
// Set time to midday to avoid dalight saving and browser quirks
s.setHours(12,0,0,0);
e.setHours(12,0,0,0);
// Get the difference in whole days
var totalDays = Math.round((e - s) / 8.64e7);
// Get the difference in whole weeks
var wholeWeeks = totalDays / 7 | 0;
// Estimate business days as number of whole weeks * 5
var days = wholeWeeks * 5;
// If not even number of weeks, calc remaining weekend days
if (totalDays % 7) {
s.setDate(s.getDate() + wholeWeeks * 7);
while (s < e) {
s.setDate(s.getDate() + 1);
// If day isn't a Sunday or Saturday, add to business days
if (s.getDay() != 0 && s.getDay() != 6) {
++days;
}
}
}
return days;
}
Dunno how it compares to jfriend00's answer or the code you referenced, if you want the period to be inclusive, just add one if the start or end date are a business day.