Hope someone can help me. User defines two dates(dd/mm/yy, then want to display 1) Difference in days between the two inputs in dd/mm/yy and 2) the difference in days between the dates. The problem is I always want to show the difference i dd/mm/yy no matter how long the period is but the maximun value I want to show for days is 1825 or 1826(which is five years depending on if it´s one or two leapyears in the given period). Please help. Give me some hlep how to do the second bit. How do I know to stop on 1825 or 1826.
Thank you so much!
Leap year(s) in five year period
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- Peter Anselmo
- Forum Commoner
- Posts: 58
- Joined: Wed Feb 27, 2008 7:22 pm
Re: Leap year(s) in five year period
I'm a little hazy on your question, I think timestamps will help you out. Use the mktime() function to convert both dates to unix timestamps (which takes into account leap years). Then find the difference, and divide by the number of seconds in a day.
Re: Leap year(s) in five year period
date('L', mktime(12, 0, 0, 1,1,'2008'));
$startYear = substr($_POST['from_yr'], -2);
$endYear = substr($_POST['to_yr'], -2);
$leaps = 0;
for($i=$startYear; $i<=$endYear; $i++) {
if(date('L', mktime(12,0,0,1,1,$i))) $leaps++;
}
If i input 28 february 2008 - 28 february 2013 i get 5 years 0 monts and 0 days AND 1827 days and one leap years.
If i input 3 march 2008 - 3 march 2013 i get 5 years 0 monts and 0 days AND 1826 days put only one leap year. <-- Here I only want it to get one leap year.
Please help.
Thank you!
$startYear = substr($_POST['from_yr'], -2);
$endYear = substr($_POST['to_yr'], -2);
$leaps = 0;
for($i=$startYear; $i<=$endYear; $i++) {
if(date('L', mktime(12,0,0,1,1,$i))) $leaps++;
}
If i input 28 february 2008 - 28 february 2013 i get 5 years 0 monts and 0 days AND 1827 days and one leap years.
If i input 3 march 2008 - 3 march 2013 i get 5 years 0 monts and 0 days AND 1826 days put only one leap year. <-- Here I only want it to get one leap year.
Please help.
Thank you!
Re: Leap year(s) in five year period
A quick search for date_diff on php.net: http://us2.php.net/manual/en/ref.datetime.php#80866
Although I would say the best solution is to use MySQL's built in datediff: http://dev.mysql.com/doc/refman/5.0/en/ ... n_datediff
Although I would say the best solution is to use MySQL's built in datediff: http://dev.mysql.com/doc/refman/5.0/en/ ... n_datediff