将一天添加到日期的代码返回日期之前的日期: 2009-09-30 20:24:00日期后增加一天应滚动到下个月:1970-01-01 17:33:29
<?php
//add day to date test for month roll over
$stop_date = date('Y-m-d H:i:s', strtotime("2009-09-30 20:24:00"));
echo 'date before day adding: '.$stop_date;
$stop_date = date('Y-m-d H:i:s', strtotime('+1 day', $stop_date));
echo ' date after adding one day. SHOULD be rolled over to the next month: '.$stop_date;
?>
我之前用过类似的代码，这里我做错了什么?
当前回答
$date = new DateTime('2000-12-31');
$date->modify('+1 day');
echo $date->format('Y-m-d') . "\n";
其他回答
$date = new DateTime('2000-12-31');
$date->modify('+1 day');
echo $date->format('Y-m-d') . "\n";
我总是加86400秒(一天):
$stop_date = date('Y-m-d H:i:s', strtotime("2009-09-30 20:24:00") + 86400);
echo 'date after adding 1 day: '.$stop_date;
这可能不是你能做到的最巧妙的方法，但它是有效的!
试试这个
echo date('Y-m-d H:i:s',date(strtotime("+1 day", strtotime("2009-09-30 20:24:00"))));
这招对我很管用: 当前日期
$date = date('Y-m-d', strtotime("+1 day"));
anydate:
date('Y-m-d', strtotime("+1 day", strtotime($date)));
简单的阅读和理解方式:
$original_date = "2009-09-29";
$time_original = strtotime($original_date);
$time_add = $time_original + (3600*24); //add seconds of one day
$new_date = date("Y-m-d", $time_add);
echo $new_date;
