Leap Years

latemodern

Hi,

My script needs to know when a date that is passed is a leap year. In the script I am listing all the leap years as:

(2008 || 2012 || 2016 || 2020 || 2024 || 2030 || 2034 || 2038 || 2042)

Is there a shorter way of writing this?

Thanks

Chris

Shrike

date( 'L', $year ) is true if $year is a leap year.

jkurrle

My way isn't as elegant, but a different option might be like this:

if (($year%4==0 && $year%100<>0)||($year%4==0 && $year%400==0))
  echo "$year is a leap year";
else
  echo "$year isn't a leap year";

Basically the rule is if the year is a leap year if:
divisible by 4
not divisible by 100
* unless the year is divisible by 400

Not that we have to worry about the 400 rule for another 394 years...

laserlight

heh, but jkurrle, if an integer is divisible by 400, it must be divisible by 4, so the test should be:

if ($year % 4 == 0 && $year % 100 != 0 || $year % 400 == 0)
	//leap year
else
	//common year

jkurrle

Would you not need to do some more grouping with parentheses? Something more like:

if ( ($year % 4 == 0 && $year % 100 != 0) || $year % 400 == 0) 
  //Leap year
else
  //Not a leap year

I'd think that otherwise there could be some confusion with the results...

laserlight

I'd think that otherwise there could be some confusion with the results...

That's true, I did deviate from good practice. The interpreter wont be confused of course, since the precedence of && is higher than that of ||

RyanJones

You can also use the date function with the L attribute... 🙂

Cheers,

Ryan Jones

laserlight

You can also use the date function with the L attribute...

That was mentioned in Shrike's reply, and it isnt so much an attribute as a format specifier.

RyanJones

laserlight wrote:
That was mentioned in Shrike's reply, and it isnt so much an attribute as a format specifier.

Thats true, an attribute is something different 🙂

Cheers,

Ryan Jones

dream_scape

Shrike wrote:
date( 'L', $year ) is true if $year is a leap year.

That won't quite work. [man]date[/man] takes a timestamp, not a string.

For it to work correctly, you have to do something like:

date('L', strtotime($year . '-01-01'));

Shrike

I don't think I elaborated on what type $year was at all, string nor timestamp. But I guess it could be construed that way 🙂

dream_scape

Shrike wrote:
I don't think I elaborated on what type $year was at all, string nor timestamp. But I guess it could be construed that way 🙂

nice try 😉

bradgrafelman

dream.scape wrote:
...For it to work correctly, you have to do something like:
date('L', strtotime($year . '-01-01'));

This is what I would use. 😉 More info on the date formats and what they return can be found on the man page for [man]date/man.

laserlight

I don't think I elaborated on what type $year was at all, string nor timestamp. But I guess it could be construed that way

I do suppose that if latemodern didnt know about date() he/she is also unlikely to know how to use the year to generate the appropriate timestamp, but then I should have noticed that your $year variable presents different semantics from jkurrle's $year variable.

I wonder if latemodern will ever read this thread again....

Shrike

If we keep making new posts he might at least get bombarded by emails 😉