你可以为下个月的第一天创建一个日期，然后使用strtotime("-1 day"， $firstOfNextMonth)

T返回给定日期当月的天数(date参见文档):

$a_date = "2009-11-23";
echo date("Y-m-t", strtotime($a_date));

这是一个完整的函数:

public function get_number_of_days_in_month($month, $year) {
    // Using first day of the month, it doesn't really matter
    $date = $year."-".$month."-1";
    return date("t", strtotime($date));
}

这将输出如下:

echo get_number_of_days_in_month(2,2014);

输出:28

这是一个更优雅的表达月底的方式:

  $thedate = Date('m/d/Y'); 
  $lastDayOfMOnth = date('d', mktime(0,0,0, date('m', strtotime($thedate))+1, 0, date('Y', strtotime($thedate)))); 

还有一个内置的PHP函数cal_days_in_month()?

"此函数将返回指定日历的某月中的天数。" http://php.net/manual/en/function.cal-days-in-month。

echo cal_days_in_month(CAL_GREGORIAN, 11, 2009); 
// = 30

这应该可以工作:

$week_start = strtotime('last Sunday', time());
$week_end = strtotime('next Sunday', time());

$month_start = strtotime('first day of this month', time());
$month_end = strtotime('last day of this month', time());

$year_start = strtotime('first day of January', time());
$year_end = strtotime('last day of December', time());

echo date('D, M jS Y', $week_start).'<br/>';
echo date('D, M jS Y', $week_end).'<br/>';

echo date('D, M jS Y', $month_start).'<br/>';
echo date('D, M jS Y', $month_end).'<br/>';

echo date('D, M jS Y', $year_start).'<br/>';
echo date('D, M jS Y', $year_end).'<br/>';