我正在尝试建立一个考勤系统。当试图得到一个小时的报告，由一个用户为给定的一天，我需要工作的总小时数。我有一个startTime(时间戳)和endTime(时间戳)，但我可以有多个条目的用户在任何给定的一天，他们在时钟去打破时钟然后时钟后和出去吃午饭等。
array
(
[theUser] => Bob Johnson
[timeId] => 9
[user_id] => 1
[weekNo] => 34
[clockYear] => 2016
[entryDate] => 2016-08-21
[startTime] => 2016-08-21 21:57:01
[endTime] => 2016-08-21 22:45:15
[updated_at] => 2016-08-21 22:45:15
[totTime] => 00:48:14
)
[9] => Array
(
[theUser] => Bob Johnson
[timeId] => 16
[user_id] => 1
[weekNo] => 34
[clockYear] => 2016
[entryDate] => 2016-08-24
[startTime] => 2016-08-24 01:00:00
[endTime] => 2016-08-24 05:15:00
[updated_at] =>
[totTime] => 04:15:00
)
(
[theUser] => Bob Johson
[timeId] => 15
[user_id] => 1
[weekNo] => 34
[clockYear] => 2016
[entryDate] => 2016-08-24
[startTime] => 2016-08-24 01:00:00
[endTime] => 2016-08-24 05:15:00
[updated_at] =>
[totTime] => 04:15:00
)
(
[theUser] => Bob Johnson
[timeId] => 18
[user_id] => 1
[weekNo] => 34
[clockYear] => 2016
[entryDate] => 2016-08-24
[startTime] => 2016-08-24 09:59:00
[endTime] => 2016-08-24 10:45:00
[updated_at] => 2016-08-24 10:19:31
[totTime] => 00:46:00
)
我需要用总工作时间来创建一个新的数组，我想实现这样的东西
(
[theUser] = Bob Johnson
[weekNo] = 34
[Year] = 2016
[entryDate] = 2016-08-24
[totalHoursWorked] = 09:16:00
[overTime] = 01:16:00
)
如果我试着服用
，有人能帮我吗? $startTime = strtotime($val['startTime']);
$endTime = strtotime($val['endTime']);
$tot += $endTime - $startTime;
echo date('H:i:s', $tot)
在foreach循环中但是我总是得到错误的值
问题出在你的最后一行:echo date('H:i:s', $tot)
你应该用gmdate代替:
echo gmdate('H:i:s', $tot);
或者用简单的数学代码格式化:
$hours = floor($init / 3600);
$minutes = floor(($init / 60) % 60);
$seconds = $init % 60;
echo "$hours:$minutes:$seconds";
使用诸如Carbon之类的库将更方便您的计算和格式化