如何在PHP中计算两个日期时间之间的分钟差异?
当前回答
另一种以分钟为单位计算差异的简单方法。请注意,这是一个在1年范围内计算的样本。详情请点击这里
$origin = new DateTime('2021-02-10 09:46:32');
$target = new DateTime('2021-02-11 09:46:32');
$interval = $origin->diff($target);
echo (($interval->format('%d')*24) + $interval->format('%h'))*60; //1440 (difference in minutes)
其他回答
时区的另一种方法。
$start_date = new DateTime("2013-12-24 06:00:00",new DateTimeZone('Pacific/Nauru'));
$end_date = new DateTime("2013-12-24 06:45:00", new DateTimeZone('Pacific/Nauru'));
$interval = $start_date->diff($end_date);
$hours = $interval->format('%h');
$minutes = $interval->format('%i');
echo 'Diff. in minutes is: '.($hours * 60 + $minutes);
function date_getFullTimeDifference( $start, $end )
{
$uts['start'] = strtotime( $start );
$uts['end'] = strtotime( $end );
if( $uts['start']!==-1 && $uts['end']!==-1 )
{
if( $uts['end'] >= $uts['start'] )
{
$diff = $uts['end'] - $uts['start'];
if( $years=intval((floor($diff/31104000))) )
$diff = $diff % 31104000;
if( $months=intval((floor($diff/2592000))) )
$diff = $diff % 2592000;
if( $days=intval((floor($diff/86400))) )
$diff = $diff % 86400;
if( $hours=intval((floor($diff/3600))) )
$diff = $diff % 3600;
if( $minutes=intval((floor($diff/60))) )
$diff = $diff % 60;
$diff = intval( $diff );
return( array('years'=>$years,'months'=>$months,'days'=>$days, 'hours'=>$hours, 'minutes'=>$minutes, 'seconds'=>$diff) );
}
else
{
echo "Ending date/time is earlier than the start date/time";
}
}
else
{
echo "Invalid date/time data detected";
}
}
<?php
$date1 = time();
sleep(2000);
$date2 = time();
$mins = ($date2 - $date1) / 60;
echo $mins;
?>
这就是我如何显示“xx次前”在php > 5.2 ..这里是DateTime对象的更多信息
//Usage:
$pubDate = $row['rssfeed']['pubDates']; // e.g. this could be like 'Sun, 10 Nov 2013 14:26:00 GMT'
$diff = ago($pubDate); // output: 23 hrs ago
// Return the value of time different in "xx times ago" format
function ago($timestamp)
{
$today = new DateTime(date('y-m-d h:i:s')); // [2]
//$thatDay = new DateTime('Sun, 10 Nov 2013 14:26:00 GMT');
$thatDay = new DateTime($timestamp);
$dt = $today->diff($thatDay);
if ($dt->y > 0){
$number = $dt->y;
$unit = "year";
} else if ($dt->m > 0) {
$number = $dt->m;
$unit = "month";
} else if ($dt->d > 0) {
$number = $dt->d;
$unit = "day";
} else if ($dt->h > 0) {
$number = $dt->h;
$unit = "hour";
} else if ($dt->i > 0) {
$number = $dt->i;
$unit = "minute";
} else if ($dt->s > 0) {
$number = $dt->s;
$unit = "second";
}
$unit .= $number > 1 ? "s" : "";
$ret = $number." ".$unit." "."ago";
return $ret;
}
以下是答案:
$to_time = strtotime("2008-12-13 10:42:00");
$from_time = strtotime("2008-12-13 10:21:00");
echo round(abs($to_time - $from_time) / 60,2). " minute";
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