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
另一种使用mktime而不是date('t')的方法:
$dateStart= date("Y-m-d", mktime(0, 0, 0, 10, 1, 2016)); //2016-10-01
$dateEnd = date("Y-m-d", mktime(0, 0, 0, 11, 0, 2016)); //This will return the last day of october, 2016-10-31 :)
所以用这种方法,它会计算它是31 30还是29
其他回答
这是一个完整的函数:
public function get_number_of_days_in_month($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
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));
如果你想回到几个月前,你也可以这样做。
$list = [
0, 1, 2, 3
];
$date = new \Datetime();
$dates = [];
foreach($list as $item)
{
$set = clone $date;
$set->modify("-$item month ");
$dates[] = $set->modify("last day of this month");
}
return $dates;
您也可以将它与datetime一起使用
$date = new \DateTime();
$nbrDay = $date->format('t');
$lastDay = $date->format('Y-m-t');
$date1 = $year.'-'.$month;
$d = date_create_from_format('Y-m',$date1);
$last_day = date_format($d, 't');