如何在PHP中获得每月的最后一天?
考虑到:
$a_date = "2009-11-23"
我想要2009-11-30;鉴于
$a_date = "2009-12-23"
我要2009-12-31。
如何在PHP中获得每月的最后一天?
考虑到:
$a_date = "2009-11-23"
我想要2009-11-30;鉴于
$a_date = "2009-12-23"
我要2009-12-31。
当前回答
function first_last_day($string, $first_last, $format) {
$result = strtotime($string);
$year = date('Y',$result);
$month = date('m',$result);
$result = strtotime("{$year}-{$month}-01");
if ($first_last == 'last'){$result = strtotime('-1 second', strtotime('+1 month', $result)); }
if ($format == 'unix'){return $result; }
if ($format == 'standard'){return date('Y-m-d', $result); }
}
http://zkinformer.com/?p=134
其他回答
现在,如果你有月份和年份,DateTime可以很方便地做到这一点
$date = new DateTime('last day of '.$year.'-'.$month);
来自另一个DateTime对象
$date = new DateTime('last day of '.$otherdate->format('Y-m'));
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/>';
试试这个,如果你使用的是PHP 5.3+,
$a_date = "2009-11-23";
$date = new DateTime($a_date);
$date->modify('last day of this month');
echo $date->format('Y-m-d');
为了查找下个月最后的日期,修改如下:
$date->modify('last day of 1 month');
echo $date->format('Y-m-d');
等等。
2行代码,你就完成了:
$oDate = new DateTime("2019-11-23");
// now your date object has been updated with last day of month
$oDate->setDate($oDate->format("Y"),$oDate->format("m"),$oDate->format("t"));
// or to just echo you can skip the above line using this
echo $oDate->format("Y-m-t");
我使用strtotime与cal_days_in_month如下所示:
$date_at_last_of_month=date('Y-m-d', strtotime('2020-4-1
+'.(cal_days_in_month(CAL_GREGORIAN,4,2020)-1).' day'));