我想从日期添加和删除一年
我有这段代码:
$yearbefore = date("Y",strtotime("-1 year"));
$yearnext = date("Y",strtotime("+1 year"));
这将从当前日期添加和删除一年,但如果我想在另一年中使用var?
$year = "2010";
如何做到这一点?
答案 0 :(得分:1)
您也可以使用strtotime
,但默认情况下需要设置日期和月份 - 因为您只需要应该没有问题的年份。
$year = intval($year); //make sure it is an integer
$yearbefore = date("Y", strtotime("01/01/$year -1 year"));
$yearafter = date("Y", strtotime("01/01/$year +1 year"));
但是当你确定,给定的年份已经是正确的格式,你可以减少/增加它。
$year = intval($year);
$yearbefore = $year - 1;
$yearafter = $year + 1;
答案 1 :(得分:0)
$date = '2013-11-01';
$seconds_in_year = 31536000;
echo date('y-m-d', strtotime($date) - $seconds_in_year);