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";
下面的代码使用DateTime类及其方法modify()和format()获得当年1月的第一天(但可以是另一个日期)，并在这一天上添加365天(但可以是N天):
echo (new DateTime((new DateTime())->modify('first day of January this year')->format('Y-m-d')))->modify('+365 days')->format('Y-m-d');
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;
由于我们经常通过API从另一个时区接收ISO字符串，我们可以动态地将其转换为本地时间:
// The machine local time is GMT+2, but we received a date at GMT+0 (UTC)
echo date(DATE_ISO8601, strtotime("-1 hour", strtotime("2022-08-17T23:25:51-00:00")));
// 2022-08-18T00:25:51+0200
最简单的解决方案:
$date = new DateTime('+1 day');
echo $date->format('Y-m-d H:i:s');
