以天为单位获取PHP日期时间差,将午夜视为一天的变化
将午夜视为一天的变化(就像以天为单位获取PHP日期时间差,将午夜视为一天的变化,php,date,datetime,dateinterval,Php,Date,Datetime,Dateinterval,将午夜视为一天的变化(就像DATEDIFF(day)SQL函数一样),获取两个PHPDateTimes之间天数差异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 解决此
DATEDIFF(day)
SQL函数一样),获取两个PHPDateTimes
之间天数差异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
解决此问题的一个简单方法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
时间在这里并不重要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
此示例可以帮助您:
$date1 = date_create($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;
请注意,DateInterval->days始终为正值。因此使用->反转
/**
* 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
}
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)
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") . ($n<0?" ago":" from now");
}
}
函数getRelativeDay(\DateTime$dt2){
如果(!$dt2){
返回false;
}
$n=getNightsBetween(new\DateTime(),$dt2);
交换机(n美元){
案例0:返回“今天”;
案例1:返回“明天”;
案例1:返回“昨天”;
违约:
返回$n.(绝对值($n)>1?“天”:“天”)($n
为什么不抓取日期部分,没有时间,并从中得到区别?如果$DATE1月和$DATE2A月份在不同的月份,这将不起作用。我怎样才能得到<代码>小时<代码>差异?这个差异方法会返回一个数字还是一个字符串?我处于类似的情况下,我需要使用一些计算的差异。再见,我在这里找到的
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");
}
}
$date1 = new DateTime("2013-08-07 13:00:00");
$date2 = new DateTime("2013-08-08 12:00:00");
$date1->setTime(0, 0, 0);
$date2->setTime(0, 0, 0);
echo $date1->diff($date2)->days;