计算工作时间，包括PHP中的制动器

Question

我正在努力编写一个PHP函数，该功能可以计算两个小时之间的时间差(减去制动器(，结果将以十进制格式。我的输入是24小时格式的字符串(HH：mm(：

$start = '07:00'; //started at 7 after midnight
$brake = '01:30'; //1 hour and 30 minutes of brake
$finish = '15:00'; //finished at 3 afternoon
//the desired result is to print out '6.5'

示例2

$start = '19:00'; //started late afternoon
$brake = '00:30'; //30 minutes of brake
$finish = '03:00'; //finished at 3 after midnight
//the desired result is to print out '7.5'

我曾经在MS Excel中具有以下公式，它效果很好：

=IF(D12>=F12,((F12+1)-D12-E12)*24,(F12-D12-E12)*24) '7.5 worked hours

其中
D12-开始时间'19：00
F12-完成时间'03：00
E12-制动时间'00：30

我试图与strtotime((一起玩，没有运气。我的PHP版本是5.4.45。请帮助

Answer 1

提供不需要时间值数学或解析的解决方案。

假设这一天尚不清楚，我们还可以考虑完成时间和开始时间的偏移，当时开始时间迟到。

示例：https：//3v4l.org/fsrbt

$start = '07:00'; //started at 7 after midnight
$break = '01:30'; //1 hour and 30 minutes of brake
$finish = '15:00'; //finished at 3 afternoon
//create the start and end date objects
$startDate = DateTime::createFromFormat('H:i', $start);
$endDate =  DateTime::createFromFormat('H:i', $finish);
if ($endDate < $startDate) {
    //end date is in the past, adjust to the next day
    //this is only needed since the day the time was worked is not known
    $endDate->add(new DateInterval('PT24H'));
}
//determine the number of hours and minutes during the break
$breakPeriod = new DateInterval(vsprintf('PT%sH%sM', explode(':', $break)));
//increase the start date by the amount of time taken during the break period
$startDate->add($breakPeriod);
//determine how many minutes are between the start and end dates
$minutes = new DateInterval('PT1M');
$datePeriods = new DatePeriod($startDate, $minutes, $endDate);
//count the number of minute date periods
$minutesWorked = iterator_count($datePeriods);
//divide the number of minutes worked by 60 to display the fractional hours
var_dump($minutesWorked / 60); //6.5

这将在24小时内00:00 - 23:59内与任何时间值一起使用。如果已知时代的一天，可以修改脚本以允许给予一天并提供更精确的时机。

Answer 2

要执行此操作，将字符串时间转换为Unix时间戳。这是自Unix时期以来的整数秒( 00:00:00协调的通用时间(UTC(，1970年1月1日，星期四，自那时以来发生的LEAP秒数(。做数学

<?php
$start = '19:00'; //started late afternoon
$break = '00:30'; //30 minutes of brake
$finish = '03:00'; //finished at 3 after midnight
//get the number of seconds for which we took a $break
//do this by converting break to unix timestamp, then extracting the hour and multiplying by 360
//and do the same extracting minutes and multiplying by 60
$breaktime = date("G",strtotime($break))*60*60 + date("i",strtotime($break))*60;
//get start time
$unixstart=strtotime($start);
//get finish time. Add a day if finish is tomorrow
if (strtotime($finish) < $unixstart) {
    $unixfinish = strtotime('+1 day', strtotime($finish));
} else {
    $unixfinish = strtotime($finish);
}
//figure out time worked
$timeworked = ($unixfinish - $unixstart - $breaktime) / 3600;
echo $timeworked;
?>

Answer 3

使用DateTime。基本上，在开始和终点时创建2个DateTime对象。到开始时间，减去制动时间，并从结果中减去结束时间。

您需要分开制动时间才能使用modify()。

<?php
$start = '07:00'; //started at 7 after midnight
$brake = '01:30'; //1 hour and 30 minutes of brake
$brakeBits = explode(":", $brake);
$finish = '15:00'; //finished at 3 afternoon
$startDate = DateTime::createFromFormat("!H:i", $start);
$startDate->modify($brakeBits[0]." hour ".$brakeBits[1]." minutes");
$endDate  = DateTime::createFromFormat("!H:i", $finish);
$diff = $startDate->diff($endDate);
echo $diff->format("%r%H:%I"); // 06:30

demo