$startDate = '2011-12-01';
    $endDate = date('Y-m');
    while (true) {
        try {
            $startDateTime = new DateTime($startDate);
            $startDateTime->add(new DateInterval('P1M'));
            $startDate = $startDateTime->format('Y-m-d');
            $endTime = $startDateTime->format('Y-m-t');
            echo $startDate . ' => ' . $endTime . PHP_EOL;
            if ($startDateTime->format('Y-m') == $endDate) {
                break;
            }
        } catch (Exception $exception) {
            var_dump($exception->getMessage());
            break;
        }
    }

在测试了许多解决方案之后，这个方法最适合我。

你的解决方案在这里。

$lastday = date('t',strtotime('today'));

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)))); 

    $startDate = '2011-12-01';
    $endDate = date('Y-m');
    while (true) {
        try {
            $startDateTime = new DateTime($startDate);
            $startDateTime->add(new DateInterval('P1M'));
            $startDate = $startDateTime->format('Y-m-d');
            $endTime = $startDateTime->format('Y-m-t');
            echo $startDate . ' => ' . $endTime . PHP_EOL;
            if ($startDateTime->format('Y-m') == $endDate) {
                break;
            }
        } catch (Exception $exception) {
            var_dump($exception->getMessage());
            break;
        }
    }

在测试了许多解决方案之后，这个方法最适合我。

$date1 = $year.'-'.$month; 
$d = date_create_from_format('Y-m',$date1); 
$last_day = date_format($d, 't');