考虑到午夜是一天的变化(就像DATEDIFF(DAY)
DateTimes
函数一样),那么获取两个PHP之间的天数差的最简单方法是什么?
例如,从今天的13:00到明天的12:00,我应该得到1(天),即使间隔不到24小时。
$date1 = new DateTime("2013-08-07 13:00:00");
$date2 = new DateTime("2013-08-08 12:00:00");
echo $date1->diff($date2)->days; // 0
发布于 2013-08-07 19:53:06
您可以忽略日期字符串的时间部分
$date1 = new DateTime(date('Y-m-d', strtotime("2013-08-07 13:00:00")));
$date2 = new DateTime(date('Y-m-d', strtotime("2013-08-08 12:00:00")));
echo $date1->diff($date2)->days; // 1
发布于 2013-08-07 19:51:09
一个简单的解决方案是剥离时间或将其设置为00:00:00
,这将始终为您提供所需的结果:
$date1 = new DateTime("2013-08-07");
$date2 = new DateTime("2013-08-08");
echo $date1->diff($date2)->days;
或
$date1 = new DateTime("2013-08-07 00:00:00");
$date2 = new DateTime("2013-08-08 00:00:00");
echo $date1->diff($date2)->days;
时间在这里并不重要
发布于 2014-04-09 23:49:42
请注意,DateInterval->days始终为正。因此使用了->invert。
/**
* return amount of days between dt1 and dt2
* (how many midnights pass going from dt1 to dt2)
* 0 = same day,
* -1 = dt2 is 1 day before dt1,
* 1 = dt2 is 1 day after dt1, etc.
*
* @param \DateTime $dt1
* @param \DateTime $dt2
* @return int|false
*/
function getNightsBetween(\DateTime $dt1, \DateTime $dt2){
if(!$dt1 || !$dt2){
return false;
}
$dt1->setTime(0,0,0);
$dt2->setTime(0,0,0);
$dti = $dt1->diff($dt2); // DateInterval
return $dti->days * ( $dti->invert ? -1 : 1); // nb: ->days always positive
}
使用示例:
$dt1 = \DateTime::createFromFormat('Y-m-d', '2014-03-03' );
$dt2 = \DateTime::createFromFormat('Y-m-d', '2014-02-20' );
getNightsBetween($dt1, $dt2); // -11
$dt1 = \DateTime::createFromFormat('Y-m-d H:i:s', '2014-01-01 23:59:59' );
$dt2 = \DateTime::createFromFormat('Y-m-d H:i:s', '2014-01-02 00:00:01' );
getNightsBetween($dt1, $dt2); // 1 (only 2 seconds later, but still the next day)
$dt1 = \DateTime::createFromFormat('Y-m-d', '2014-04-09' );
$dt2 = new \DateTime();
getNightsBetween($dt1, $dt2); // xx (how many days (midnights) passed since I wrote this)
一些文本魔术的例子:
function getRelativeDay(\DateTime $dt2){
if(!$dt2){
return false;
}
$n = getNightsBetween( new \DateTime(), $dt2);
switch($n){
case 0: return "today";
case 1: return "tomorrow";
case -1: return "yesterday";
default:
return $n . (abs($n)>1?"days":"day") . ($n<0?" ago":" from now");
}
}
https://stackoverflow.com/questions/18102603
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