How to calculate leap year

mistu2020

I want to calculate no of days.

If leap year then the no of days will be 366 other wise 365.

How to do this????

mohana_cse06

If year MOD 4 = 0 then leap year

ex: 2014 MOD 4 = 2 means not a leap year
2016 MOD 4 = 0 means Leap year.

lvanvugt

Leap year algorithm

geordie

Another alternative:

NoOfDays := DMY2DATE(31,12,Year) - DMY2DATE(31,12,Year - 1);

(Simultaneous posting :oops: )

Marije_Brummel

You can also use the virtual date table. Count the days in a year. It's an in memory process, should be fast.

ta5

Just curiuos:
How about use a .NET Class for this?
Thomas

Marije_Brummel

Good thinking Thomas, we should think .net these days.

http://msdn.microsoft.com/en-us/library ... r(v=vs.110).aspx

lvanvugt

Mark Brummel wrote:

Good thinking Thomas, we should think .net these days.

:thumbsup:

ta5

Mark Brummel wrote:

Good thinking Thomas, we should think .net these days.

And IMO at least as important: We should not re-invent the wheel, this was the reason for my first answer.

Regards
Thomas

MBerger

Mark Brummel wrote:

Good thinking Thomas, we should think .net these days.

True, but in this case it seems to me that using .net to determine if a year is a leap year is a bit overkill and more using the technology for the sake of having used the technology......

It is as when the Americans met the Russians for the first time in space : The US had invested millions in making a pen with a pressurized ink cartridge so it would write in zero gravity. When they asked the russians how they solved that particular problem, the Russians answered : "we just use pencils".

[edit] To add yet another way to determine the leap year : Check the date table if February 29th exists for that year.

kriki

And this is pure Russian-style C/AL 8) :

TheYear := 2012;

Date.GET(Date."Period Type"::Month,DMY2DATE(1,2,TheYear));
IF DATE2DMY(NORMALDATE(Date."Period End"),1) = 29 THEN
  MESSAGE('Year %1 is a leap year',TheYear);

BTW: Mark, your link does not work. This is the correct link : http://msdn.microsoft.com/en-us/library/system.datetime.isleapyear(v=vs.110).aspx

beran

I have won many a beer on a debate when year is a leap year.

A leap year is a year that can be divided by 4 unless it can be divided by 100 and again if it can be divided by 400.

So 1700, 1800, 1900 was not a leap year, but 2000 was, and 2100 will not be a leap year.

And for calculation in practice then do take some .net class that can calculate it as many suggest or your AL code. If date between February 28 <year> to March 1 <year> is equal to 2 then a leap year otherwise not.

einsTeIn.NET

beran wrote:

I have won many a beer on a debate when year is a leap year.

A leap year is a year that can be divided by 4 unless it can be divided by 100 and again if it can be divided by 400.

So 1700, 1800, 1900 was not a leap year, but 2000 was, and 2100 will not be a leap year.

Me too

But I'll need to pay as much beer if my software still runs in 2100. Because year 2000 can be devided by 400 I sometimes tend to just divide by 4. I know this is not accurate but I believe it's sufficient for the scope of my code.

Here's another solution:

IF DATE2DMY(CALCDATE('<CM>',DMY2DATE(1,2,MyYear)),1) = 29 THEN
  LeapYear := TRUE;

ta5

This discussion shows (to me at least) that normally you are better off using a built in function, in this case the .NET. Of course .NET function could be wrong, but I trust it's not...
Thomas

einsTeIn.NET

To be honest, the creation of the Date table is also a built-in function. So, I would trust it also.

PerJuhl

mohana_cse06 wrote:

If year MOD 4 = 0 then leap year

ex: 2014 MOD 4 = 2 means not a leap year
2016 MOD 4 = 0 means Leap year.

This will not work, as 1700, 1800, 1900 and 2100 eg. are not leap years.

BR Per

PerJuhl

geordie wrote:

Another alternative:

NoOfDays := DMY2DATE(31,12,Year) - DMY2DATE(31,12,Year - 1);

(Simultaneous posting :oops: )

:thumbsup: