cal_days_in_month

(PHP 4 >= 4.1.0, PHP 5, PHP 7, PHP 8)

cal_days_in_monthSeçilen takvime ve yıla göre istenen aydaki gün sayısını geri döndürür

Açıklama

cal_days_in_month(int $takvim, int $ay, int $yil): int

İşlev seçilen takvim, yil ve ay bağımsız değişkenlerine göre gün sayısını geri döndürecektir.

Bağımsız Değişkenler

takvim

Hesaplama yapılacak takvim

ay

Seçilen takvimdeki ay

yil

Seçilen takvimdeki yıl

Dönen Değerler

Bağımsız değişken olarak verilen takvimde, seçilen aydaki gün sayısı

Örnekler

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";
?>

add a note

User Contributed Notes 3 notes

up
145
brian at b5media dot com
16 years ago
Remember if you just want the days in the current month, use the date function:
$days = date("t");
up
45
dbindel at austin dot rr dot com
20 years ago
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 Bindel
up
2
datlx at yahoo dot com
2 years ago
function 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;
}
}
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