cal_days_in_month
(PHP 4 >= 4.1.0, PHP 5)
cal_days_in_month — Return the number of days in a month for a given year and calendar
Description
int cal_days_in_month
( int
$calendar
, int $month
, int $year
)
This function will return the number of days in the
month of year for
the specified calendar.
Parameters
-
calendar -
Calendar to use for calculation
-
month -
Month in the selected calendar
-
year -
Year in the selected calendar
Return Values
The length in days of the selected month in the given calendar
Examples
Example #1 cal_days_in_month() example
<?php
$num = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31
echo "There was $num days in August 2003";
?>
Коментарии
Here's a one-line function I just wrote to find the numbers of days in a month that doesn't depend on any other functions.
The reason I made this is because I just found out I forgot to compile PHP with support for calendars, and a class I'm writing for my website's open source section was broken. So rather than recompiling PHP (which I will get around to tomorrow I guess), I just wrote this function which should work just as well, and will always work without the requirement of PHP's calendar extension or any other PHP functions for that matter.
I learned the days of the month using the old knuckle & inbetween knuckle method, so that should explain the mod 7 part. :)
<?php
/*
* days_in_month($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))) : (($month - 1) % 7 % 2 ? 30 : 31);
}
?>
Enjoy,
David Bindel
Remember if you just want the days in the current month, use the date function:$days = date("t");
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;
}
}