我有一个时间列$data['Time'] (hh:mm:ss)
,需要将其转换为分钟。我该怎么做?当我这样写的时候:
$avg = ($data['Kilometers'] / $data['Time']) * 60;
我有这个错误
Warning: Division by zero in ... on line ..
试试这样的东西:
function minutes($time){
$time = explode(':', $time);
return ($time[0]*60) + ($time[1]) + ($time[2]/60);
}
现场演示
$time = explode(':', $data['Time']);
$minutes = ($time[0] * 60.0 + $time[1] * 1.0);
$avg = $minutes > 0 ? $data['Kilometers'] / $minutes : 'inf'; // if time stored is 0, then average is infinite.
另一种将时间戳转换为分钟的方法是,
$time = date('i', strtotime($data['Time']));
这是一个简单的计算使用这个
$time='01:02:10';
$timesplit=explode(':',$time);
$min=($timesplit[0]*60)+($timesplit[1])+($timesplit[2]>30?1:0);
echo $min.' Min' // 62 Min
You can use following function
<?php
function date2min ($hms) {
$fromTime = strtotime($hms);
$getMins = round(abs($fromTime) / 60,2);
return $getMins;
}
$date = date('H:i:s'); // $data['Time'] your desired time
$myResult = date2min($date);
and then use $myResult value according to your need.
?>
如果hh:mm:ss是一个字符串,您可以尝试这个经过测试的字符串,它可以工作:
$hour=substr($string, 0,1);// we get the first two values from the hh:mm:ss string
$hour=(int)$hour;
$hourtomin=$hour*60;// after we have the hour we multiply by 60 to get the min
$min=substr($string, 3,4);//now we do a substring 3 to 4 because we want to get only the min, and we don't want to get the : which is in position 2
$min=(int)$min;
$totalmin=$hourtomin+$min;// we just do a simple sum to calculate all the min
echo $totalmin;
你可以用非常简单的方法。
$minutes=$item['time_diff'];
$hours = sprintf('%02d',intdiv($minutes, 60)) .':'. ( sprintf('%02d',$minutes % 60));
return $hours;
输出将09:10
function hourMinute2Minutes($strHourMinute) {
$from = date('Y-m-d 00:00:00');
$to = date('Y-m-d '.$strHourMinute.':00');
$diff = strtotime($to) - strtotime($from);
$minutes = $diff / 60;
return (int) $minutes;
}
echo hourMinute2Minutes('01:30'); // returns 90