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我想从日期中添加和删除一年

我有这个代码:

$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,但是如果我想在 var 中从另一年开始呢?

$year = "2010";

如何才能做到这一点?

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2 回答 2

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您也可以使用strtotime,但需要默认设置日期和月份 - 因为您只需要应该没问题的年份。

$year = intval($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"));

但是当您确定给定年份的格式已经正确时,您可以减少/增加它。

$year = intval($year);
$yearbefore = $year - 1;
$yearafter  = $year + 1;
于 2013-11-02T01:35:54.143 回答
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$date = '2013-11-01';
$seconds_in_year = 31536000;
echo date('y-m-d', strtotime($date) - $seconds_in_year);
于 2013-11-02T02:03:41.007 回答