calc下2日期间的工作时间在PHP

Question

我有一个函数返回两个日期之间的差别，但是我需要计算出的差异在工作时间，假设周一至周五（上午9:00至下午5:30）：

//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 3时58分58秒
日期2 = 2012-03-22 11点29分16秒

是否有这样做的一个简单的方法，即 - 计算一周内工作时间的百分比并将差值除以我们荷兰国际集团上述功能 - 我打周围的这个想法，得到了一些很奇怪的数字...

还是有更好的办法....？

Answer 1

本例中使用PHP的内置的DateTime类来完成的日期数学。我是如何处理这个问题的，首先计算这两个日期之间的全部工作日数，然后乘以8（见注释）。然后获得部分日子的工作时间，并将其添加到工作的总小时数中。把它变成一个函数将是相当直接的。

注：

• 不带时间戳考虑。但你已经知道如何做到这一点。
• 没有处理假期。 （可以通过使用一系列节假日并将其添加到星期六和星期日过滤的地方轻松添加）。
• 需要PHP 5.3.6+
• 假定一个8小时工作日。如果员工不吃午餐$hours = $days * 8;至$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; 

See it in action

参考

• DateTime Class
• DatePeriod Class
• DateInterval Class

Answer 2

下面是我想出。

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对其进行调整（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）。

在完成开始和结束时间后，比较时间以检索DateInterval差异，计算总天数，小时数等等。然后检查日期范围是否有任何周末天数，如果找到，总差额从差异中减少。

最后，作为评价的时间被计算。:)

欢呼到John鼓舞人心的一些解决方案，特别是DatePeriod检查周末。

金星给谁打破这个;如果有人发现漏洞，我会很乐意更新！

---

金星给自己，我打破了它！是的，周末仍然是越野车（尝试从周六下午4点开始，周一下午1点结束）。我将征服你，工作时间问题！

忍者编辑＃2：如果周末出现问题，我想通过将开始和结束时间恢复到最近的相应工作日来照顾周末的错误。测试了一些日期范围（如预期的那样，在同一个周末的酒吧上开始和结束），取得了很好的效果。我并不完全相信这是尽可能的优化/简单，但至少现在它效果更好。

---

// 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); 

Answer 3

正如那句老话：“如果你想要的东西做对自己动手”。不是说这是最佳的，但它至少会为我返回正确的小时数。

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;