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    cal_days_in_month

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

    cal_days_in_month — 指定した年とカレンダーについて、月の日数を返す

    説明

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

    この関数は、指定した calendar について year 年 month 月の日数を返します。

    パラメータ

    calendar

    計算に使用するカレンダー。

    month

    選択したカレンダーにおける月。

    year

    選択したカレンダーにおける年。

    戻り値

    指定したカレンダーの、その月の日数を返します。

    例

    例1 cal_days_in_month() の例

    <?php
    $number     = cal_days_in_month(CAL_GREGORIAN, 8, 2003); // 31
    echo "2003 年 8 月の日数は {$number} 日です";
    ?>

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    User Contributed Notes 3 notes

    up
    145
    brian at b5media dot com
    17 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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