algorithm - 计算午夜时差

Question

这是我永远无法工作的一件事。
我的问题是检测一天的结束和下一天的开始，然后将差异分成每一天。

假设您想计算一个工资率，但它必须跨越午夜。

它也适用于计算在定时系统上运行的时间，或者它应该运行的时间差。

Answer 1

我更喜欢在 Unix Epoch 中记录所有时间，这样就很简单

hoursWorked = ((stopTime - startTime)/60)/60

这是一个完整的解决方案，它是用 PHP 编写的，但应该很容易构建成任何语言:

<?php
$startTime = "12/31/2008 22:02"; //No AM/PM we'll use 24hour system
$endTime = "01/01/2009 06:27"; //Again no AM/PM and we spaned the midnight time gap.
/*
Use this to test for a normal shift not ocurring durring midnight.
$startTime = "01/01/2008 06:02"; //No AM/PM we'll use 24hour system
$endTime = "01/01/2008 14:27"; //Again no AM/PM and we spaned the midnight time gap.
*/
$startTime = split(' ', $startTime);
$endTime = split(' ', $endTime);
$startTime[1] = split(':', $startTime[1]);
$endTime[1] = split(':', $endTime[1]);
/*
$startTime[0] contains the date
$startTime[1][0] contains the hours
$startTime[1][1] contains the minutes
same is true for endTime
*/
if($startTime[0] != $endTime[0])
 {
    if(date2epoch($endTime[0]) > date2epoch($startTime[0]))
     {
        $minutesWorked1 = (59 - $startTime[1][1]); //Calculate how many minutes have occured from begining of shift to midnight
        $minutesWorked2 = $endTime[1][1]; //Number of minute from midnight to end of shift
        $hoursWorked1 = (23 - $startTime[1][0]);//Number of hours from start of shift to midnight
        $hoursWorked2 = $endTime[1][0];//Number of minutes from midnight to end of shift
        echo 'Before midnight you worked ' . $hoursWorked1 . ':' . $minutesWorked1 . "\nAfter midnight you worked " . $hoursWorked2 . ':' . $minutesWorked2 . '.';
     }
    else 
     {
        //SOMETHING MAJOR BAD HAS HAPPENED WHILE LOGGING THE CLOCKINS AND CLOCKOUTS
     }
 }
else 
 {
    echo 'Today you worked ' . ($endTime[1][0] - $startTime[1][0]) . ':' . ($endTime[1][1] - $startTime[1][1]);
 }
function date2epoch($date, $format='m/d/Y')
 {
    $date = split('/', $date);
    return mktime('0', '0', '0', $date[0], $date[1], $date[2]);
 }
?>