cal_days_in_month

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

cal_days_in_month — 返回指定历法中某年某月的天数

说明

cal_days_in_month(int $calendar, int $month, int $year): int

此函数返回指定 calendar 中的某 year 中的某 month 的天数。

参数

calendar: 用来计算的历法
month: 指定历法中的某月
year: 指定历法中的某年

返回值

指定历法中所选月份的天数

示例

示例 #1 cal_days_in_month() 示例

<?php
$number = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31
echo "There were {$number} days in August 2003";
?>

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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");

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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

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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;
    }
}