以天为单位获取 PHP DateTime 差异，将午夜视为一天的变化
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Get a PHP DateTime difference in days, considering midnight as a day change
提问by marcv
What is the simplest way to get the difference in days between two PHP DateTimes, considering midnight as a day change (just like the DATEDIFF(DAY)SQL function does)?
DateTimes将午夜视为一天的变化（就像DATEDIFF(DAY)SQL 函数一样），获取两个 PHP 之间的天数差异的最简单方法是什么？
For example, between today at 13:00 and tomorrow at 12:00, I should get 1 (day), even though the interval is less than 24 hours.
例如，今天13:00到明天12:00之间，即使间隔小于24小时，我也应该得到1（天）。
$date1 = new DateTime("2013-08-07 13:00:00");
$date2 = new DateTime("2013-08-08 12:00:00");
echo $date1->diff($date2)->days; // 0
回答by MajorCaiger
You could ignore the time portion of the date string
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
回答by x4rf41
a simple solution to this to strip the time or set it to 00:00:00, that should always give you the desired result:
对此的一个简单解决方案是剥离时间或将其设置为00:00:00，这应该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;
or
或者
$date1 = new DateTime("2013-08-07 00:00:00");
$date2 = new DateTime("2013-08-08 00:00:00");
echo $date1->diff($date2)->days;
the time doesnt really matter here
时间在这里并不重要
回答by MoonLite
note that DateInterval->days is always positive. hence the use of ->invert.
请注意 DateInterval->days 始终为正数。因此使用 ->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
}
usage examples:
用法示例：
$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)
example of some text magic:
一些文本魔法的例子：
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");
}
}
回答by sprutex
$date1 = new DateTime("2013-08-07 13:00:00");
$date2 = new DateTime("2013-08-08 12:00:00");
- Drop time in DateTime's objects:
- 在 DateTime 的对象中删除时间：
$date1->setTime(0, 0, 0);
$date2->setTime(0, 0, 0);
- Get difference of adapted objects:
- 获取适配对象的差异：
echo $date1->diff($date2)->days;
回答by Samira Khorshidi
this example can help you:
这个例子可以帮助你：
$date1 = date_create($d1);
$date2 = date_create($d2);
//$FromFullDateTime=$from.$FromTime;
$date1_month = date_format($date1, 'd');
$date2_month = date_format($date2, 'd');
$dif = $date2_month - $date1_month;
