Remember if you just want the days in the current month, use the date function:
$days = date("t");
(PHP 4 >= 4.1.0, PHP 5, PHP 7, PHP 8)
cal_days_in_month — Seçilen takvime ve yıla göre istenen aydaki gün sayısını geri döndürür
İşlev seçilen takvim, yil ve
ay bağımsız değişkenlerine göre gün sayısını geri
döndürecektir.
takvimHesaplama yapılacak takvim
aySeçilen takvimdeki ay
yilSeçilen takvimdeki yıl
Bağımsız değişken olarak verilen takvimde, seçilen aydaki gün sayısı
Örnek 1 cal_days_in_month() örneği
<?php
$num = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31
echo "2003 yılı Ağustos ayında {$num} gün vardır";
?>Remember if you just want the days in the current month, use the date function:
$days = date("t");Here's a one-line function I just wrote to find the numbers of days in a month that doesn't depend on any other functions.
The reason I made this is because I just found out I forgot to compile PHP with support for calendars, and a class I'm writing for my website's open source section was broken. So rather than recompiling PHP (which I will get around to tomorrow I guess), I just wrote this function which should work just as well, and will always work without the requirement of PHP's calendar extension or any other PHP functions for that matter.
I learned the days of the month using the old knuckle & inbetween knuckle method, so that should explain the mod 7 part. :)
<?php
/*
* days_in_month($month, $year)
* Returns the number of days in a given month and year, taking into account leap years.
*
* $month: numeric month (integers 1-12)
* $year: numeric year (any integer)
*
* Prec: $month is an integer between 1 and 12, inclusive, and $year is an integer.
* Post: none
*/
// corrected by ben at sparkyb dot net
function days_in_month($month, $year)
{
// calculate number of days in a month
return $month == 2 ? ($year % 4 ? 28 : ($year % 100 ? 29 : ($year % 400 ? 28 : 29))) : (($month - 1) % 7 % 2 ? 30 : 31);
}
?>
Enjoy,
David Bindelfunction lastDayOfMonth(string $time, int $deltaMonth, string $format = 'Y-m-d')
{
try {
$year = date('Y', strtotime($time));
$month = date('m', strtotime($time));
$increaYear = floor(($deltaMonth + $month - 1) / 12);
$year += $increaYear;
$month = (($deltaMonth + $month) % 12) ?: 12;
$day = cal_days_in_month(CAL_GREGORIAN, $month, $year);
return $time . ' + ' . $deltaMonth . ' => ' . date($format, strtotime($year . '-' . $month . '-' . $day)) . "\n";
} catch (Exception $e) {
throw $e;
}
}