4

我有一个函数可以返回两个日期之间的差异，但是我需要计算出工作时间的差异，假设星期一到星期五（上午 9 点到下午 5 点 30 分）：

//DATE DIFF FUNCTION
// Set timezone
date_default_timezone_set("GMT");

// Time format is UNIX timestamp or
// PHP strtotime compatible strings
function dateDiff($time1, $time2, $precision = 6) {
    // If not numeric then convert texts to unix timestamps
    if (!is_int($time1)) {
        $time1 = strtotime($time1);
    }
    if (!is_int($time2)) {
        $time2 = strtotime($time2);
    }

    // If time1 is bigger than time2
    // Then swap time1 and time2
    if ($time1 > $time2) {
        $ttime = $time1;
        $time1 = $time2;
        $time2 = $ttime;
    }

    // Set up intervals and diffs arrays
    $intervals = array('year','month','day','hour','minute','second');
    $diffs = array();

    // Loop thru all intervals
    foreach ($intervals as $interval) {
        // Set default diff to 0
        $diffs[$interval] = 0;
        // Create temp time from time1 and interval
        $ttime = strtotime("+1 " . $interval, $time1);
        // Loop until temp time is smaller than time2
        while ($time2 >= $ttime) {
            $time1 = $ttime;
            $diffs[$interval]++;
            // Create new temp time from time1 and interval
            $ttime = strtotime("+1 " . $interval, $time1);
        }
    }

    $count = 0;
    $times = array();
    // Loop thru all diffs
    foreach ($diffs as $interval => $value) {
        // Break if we have needed precission
        if ($count >= $precision) {
            break;
        }
        // Add value and interval 
        // if value is bigger than 0
        if ($value > 0) {
            // Add s if value is not 1
            if ($value != 1) {
                $interval .= "s";
            }
            // Add value and interval to times array
            $times[] = $value . " " . $interval;
            $count++;
        }
    }

    // Return string with times
    return implode(", ", $times);
}

日期 1 = 2012-03-24 03:58:58
日期 2 = 2012-03-22 11:29:16

有没有一种简单的方法可以做到这一点，即 - 计算一周工作时间的百分比并使用上述函数除以差异 - 我已经尝试过这个想法并得到了一些非常奇怪的数字......

还是有更好的方法……？

4

3 回答 3

4

此示例使用 PHP 内置的 DateTime 类来进行日期数学运算。我的方法是从计算两个日期之间的完整工作日数开始，然后将其乘以 8（见注释）。然后它会获取部分工作日的工作时间并将它们添加到总工作时间中。把它变成一个函数是相当简单的。

笔记：

  • 不考虑时间戳。但是您已经知道如何做到这一点。
  • 不处理假期。（可以通过使用一系列假期并将其添加到您过滤掉周六和周日的位置来轻松添加）。
  • 需要 PHP 5.3.6+
  • 假设每天工作 8 小时。如果员工不采取午餐$hours = $days * 8;更改$hours = $days * 8.5;

.

<?php
// Initial datetimes
$date1 = new DateTime('2012-03-22 11:29:16');
$date2 = new DateTime('2012-03-24 03:58:58');

// Set first datetime to midnight of next day
$start = clone $date1;
$start->modify('+1 day');
$start->modify('midnight');

// Set second datetime to midnight of that day
$end = clone $date2;
$end->modify('midnight');

// Count the number of full days between both dates
$days = 0;

// Loop through each day between two dates
$interval = new DateInterval('P1D');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $dt) {
    // If it is a weekend don't count it
    if (!in_array($dt->format('l'), array('Saturday', 'Sunday'))) {
        $days++;
    }
}

// Assume 8 hour workdays
$hours = $days * 8;

// Get the number of hours worked on the first day
$date1->modify('5:30 PM');
$diff = $date1->diff($start);
$hours += $diff->h;

// Get the number of hours worked the second day
$date1->modify('8 AM');
$diff = $date2->diff($end);
$hours += $diff->h;

echo $hours;

看到它在行动

参考

于 2013-12-19T13:56:31.453 回答
2

这就是我想出的。

我的解决方案检查原始日期的开始和结束时间，并根据工作日的实际开始和结束时间进行调整（如果原始开始时间在工作开始时间之前，则将其设置为后者）。

在对开始时间和结束时间完成此操作后，比较时间以检索DateInterval差异，计算总天数、总小时数等。然后检查日期范围的任何周末，如果找到，则从总天数减少差异。

最后，按照评论计算小时数。:)

为John提供的一些解决方案干杯，尤其是DatePeriod检查周末。

任何打破这一点的人的金星；如果有人发现漏洞，我会很乐意更新！

给自己的金星，我打破了它！是的，周末仍然有问题（尝试从周六下午 4 点开始，到周一下午 1 点结束）。我会征服你的，工作时间问题！

忍者编辑#2：我想我通过将开始和结束时间恢复到最近的相应工作日来解决周末错误，如果它们是在周末的话。在测试了几个日期范围后得到了很好的结果（如预期的那样，在同一个周末 barfs 开始和结束）。我并不完全相信这是尽可能优化/简单的，但至少现在效果更好。

// Settings
$workStartHour = 9;
$workStartMin = 0;
$workEndHour = 17;
$workEndMin = 30;
$workdayHours = 8.5;
$weekends = ['Saturday', 'Sunday'];
$hours = 0;

// Original start and end times, and their clones that we'll modify.
$originalStart = new DateTime('2012-03-22 11:29:16');
$start = clone $originalStart;

// Starting on a weekend? Skip to a weekday.
while (in_array($start->format('l'), $weekends))
{
    $start->modify('midnight tomorrow');
}

$originalEnd = new DateTime('2012-03-24 03:58:58');
$end = clone $originalEnd;

// Ending on a weekend? Go back to a weekday.
while (in_array($end->format('l'), $weekends))
{
    $end->modify('-1 day')->setTime(23, 59);
}

// Is the start date after the end date? Might happen if start and end
// are on the same weekend (whoops).
if ($start > $end) throw new Exception('Start date is AFTER end date!');

// Are the times outside of normal work hours? If so, adjust.
$startAdj = clone $start;

if ($start < $startAdj->setTime($workStartHour, $workStartMin))
{
    // Start is earlier; adjust to real start time.
    $start = $startAdj;
}
else if ($start > $startAdj->setTime($workEndHour, $workEndMin))
{
    // Start is after close of that day, move to tomorrow.
    $start = $startAdj->setTime($workStartHour, $workStartMin)->modify('+1 day');
}

$endAdj = clone $end;

if ($end > $endAdj->setTime($workEndHour, $workEndMin))
{
    // End is after; adjust to real end time.
    $end = $endAdj;
}
else if ($end < $endAdj->setTime($workStartHour, $workStartMin))
{
    // End is before start of that day, move to day before.
    $end = $endAdj->setTime($workEndHour, $workEndMin)->modify('-1 day');
}

// Calculate the difference between our modified days.
$diff = $start->diff($end);

// Go through each day using the original values, so we can check for weekends.
$period = new DatePeriod($start, new DateInterval('P1D'), $end);

foreach ($period as $day)
{
    // If it's a weekend day, take it out of our total days in the diff.
    if (in_array($day->format('l'), ['Saturday', 'Sunday'])) $diff->d--;
}

// Calculate! Days * Hours in a day + hours + minutes converted to hours.
$hours = ($diff->d * $workdayHours) + $diff->h + round($diff->i / 60, 2);
于 2013-12-20T04:44:56.797 回答
1

俗话说“如果你想把事情做好，就自己做”。并不是说这是最佳的，但它至少为我返回了正确的小时数。

function biss_hours($start, $end){

    $startDate = new DateTime($start);
    $endDate = new DateTime($end);
    $periodInterval = new DateInterval( "PT1H" );

    $period = new DatePeriod( $startDate, $periodInterval, $endDate );
    $count = 0;

      foreach($period as $date){

           $startofday = clone $date;
           $startofday->setTime(8,30);

           $endofday = clone $date;
           $endofday->setTime(17,30);

    if($date > $startofday && $date <= $endofday && !in_array($date->format('l'), array('Sunday','Saturday'))){

        $count++;
    }

}

//Get seconds of Start time
$start_d = date("Y-m-d H:00:00", strtotime($start));
$start_d_seconds = strtotime($start_d);
$start_t_seconds = strtotime($start);
$start_seconds = $start_t_seconds - $start_d_seconds;

//Get seconds of End time
$end_d = date("Y-m-d H:00:00", strtotime($end));
$end_d_seconds = strtotime($end_d);
$end_t_seconds = strtotime($end);
$end_seconds = $end_t_seconds - $end_d_seconds;

$diff = $end_seconds-$start_seconds;

if($diff!=0):
    $count--;
endif;

$total_min_sec = date('i:s',$diff);

return $count .":".$total_min_sec;
}

$start = '2014-06-23 12:30:00';
$end = '2014-06-27 15:45:00';

$go = biss_hours($start,$end);
echo $go;
于 2014-07-01T09:30:57.163 回答