$ymr='2005-06-09';
$mr=substr($ymr,4);
$ymr= strtotime($ymr)+366*24*60*60;
$y=strftime('%Y',$ymr);
$ymr=$y.$mr;
$mr=substr($ymr,4);
$ymr= strtotime($ymr)+366*24*60*60;
$y=strftime('%Y',$ymr);
$ymr=$y.$mr;
$ymr= strtotime($ymr)+366*24*60*60;
else
$ymr= strtotime($ymr)+365*24*60*60;$ymr=strftime('%Y-%m-%d',$ymr);
改用这个
$ymr='2005-06-09';
$m=substr($ymr,5,2);
$y=substr($ymr,0,4);if($m>2)
$y=$y+1;
if(checkdate('2','29',$y)) $ymr= strtotime($ymr)+366*24*60*60;else
$ymr= strtotime($ymr)+365*24*60*60;$ymr=strftime('%Y-%m-%d',$ymr);这个例子算出的结果和下面的是一样的:$ymr='2005-06-09';
$mr=substr($ymr,4);
$ymr= strtotime($ymr)+366*24*60*60;
$y=strftime('%Y',$ymr);
$ymr=$y.$mr;
加一年就是加一年,不需要考虑润年问题php
echo date("Y-m-d H:i:s",strtotime("2005-06-08 11:18:00 +1 year")); //out 2006-06-08 11:18:00mysql
select adddate('2005-06-08 11:18:00', interval 1 year)
out:
2006-06-08 11:18:00