JavaScript function that accepts a number from 1 - 12 representing the months of the year, and then returns the number of days in that month?
+5
A:
Try this:
function numberOfDays(year, month) {
var d = new Date(year, month, 0);
return d.getDate();
}
Because of leap years you need to pass the year too.
Caleb
2009-11-27 23:29:17
This is almost correct except that as far as Javascript's `Date` objects are concerned, months are indexed from 0-11 and the OP wants to input months indexed from 1-12. You'll have to subtract 1 from the month variable thats passed in if you want my upvote.
Asaph
2009-11-28 00:58:56
@Asaph: look at the answer again. It's perfect.
Crescent Fresh
2009-11-28 01:43:32
@Crescent Fresh: Ok, I see what's going on now. The month needs 1 subtracted from it to correct for the zero indexing but then the strategy used is to examine the next month (ie. add 1 to the month, thereby canceling out the subtracted 1) and back off the day (3rd argument to `Date` constructor) by 1, making it the zeroth day of next month which works out to the last day of the current month. So it works. But it's clever. And clever code is unmaintainable code. I would refactor this to something more readable or at the very least add a comment to explain the cleverness.
Asaph
2009-11-28 03:00:44
@Asaph: "clever" is subjective. The algorithm is sound. About the only negative IMO is it does hit the system clock to do something that can be determined with simple conditionals and math.
Crescent Fresh
2009-11-28 04:02:14
+7
A:
function getDaysInMonth(m, y) {
return /8|3|5|10/.test(--m)?30:m==1?(!(y%4)&&y%100)||!(y%400)?29:28:31;
}
J-P
2009-11-27 23:32:57
+3
A:
Thirty days hath September,
April, June, and November;
All the rest have thirty-one,
Excepting February alone,
Which hath twenty-eight days clear,
And twenty-nine in each leap year.
James
2009-11-27 23:37:28
A:
try this:
function DaysinMonth(aDate) {
return aDate.setMonth(aDate.getMonth()+1, 0).getDate();
}
Charles Bretana
2009-11-27 23:55:22
Why would you mutate the argument passed in? That's childish man, childish.
Crescent Fresh
2009-11-28 01:50:19
+1
A:
In the spirit of not doing your homework for you, I present a version in POVRay (sorry, not JS) I did many years ago.
In POVRay, there are no boolean variables. The method I came up with was to create a polynomial in 'm' which gave an answer > 0 for months with 31 days and < 0 for months with 30 days.
#declare m0 = (m-0.5)*(m-1.5)*(m-2.5)*(m-3.5)*(m-4.5)*(m-5.5);
#declare m0 = m0*(m-6.5)*(m-8.5)*(m-9.5)*(m-10.5)*(m-11.5);
#if (m0 > 0)
#declare maxdays = 31;
#else
#declare maxdays = 30;
#end
The tricky part is to decide when the year is a leap year. This is the full test for leap years. Most people are aware of the 4-year rule, and since 2000, some know about the 100 and 400 year rules, there is no 4000 year rule.
#declare LEAPYEAR = 2.0;
#if (mod(YEAR,4.0)=0)
#declare LEAPYEAR = 1.0;
#if (mod(YEAR,100.0)=0)
#declare LEAPYEAR = 2.0;
#end
#if (mod(YEAR,400.0)=0
#declare LEAPYEAR = 1.0;
#end
#end
#if (MONTH = 2.0)
#declare maxdays = maxdays - LEAPYEAR;
#end
#if (DAY > maxdays)
#declare MONTH = MONTH + 1;
#declare DAY = DAY - maxdays;
#end
#if (MONTH > 12)
#declare YEAR = YEAR + 1;
#declare MONTH = MONTH - 12;
#end
pavium
2009-11-28 00:18:52