我想计算两个日期之间的工作时间。所以我需要忽略周六/周日，我的工作时间是上午 9 点到下午 6 点（9:00 到 18:00）。
我在网上搜索并尝试了一些变化，但似乎没有任何效果。任何人都可以帮忙吗？
我一直在看这个链接： 计算两个日期之间的 工作时间从 php 中的两个给定时间减去非工作时间和天数 计算 PHP 中两个日期之间的工作时间
但是，或者他们没有得到答案，或者答案提供了它不起作用。
谢谢你的帮助
问问题
3373 次
3 回答
2
我根据自己的需要在 web 和 stackoverflow 上调整了一些功能。他们来了。
调用脚本：
date_default_timezone_set("Europe/Lisbon");
$startDate=strtotime('2012-11-30 00:15:33');
$endDate = strtotime('2012-12-05 10:15:00');
echo "Working hours from create date until due date -> ".seconds2human(work_hours_diff($startDate,$endDate));
function seconds2human($ss) {
$s = $ss%60;
$m = floor(($ss%3600)/60);
$h = floor(($ss)/3600);
return "$h hours, $m minutes, $s seconds";
}
使用工作日和工作时间的子跟踪日期（改编自计算两个日期之间的工作时间）：
function work_hours_diff($date1,$date2) {
if ($date1>$date2) {
$tmp=$date1;
$date1=$date2;
$date2=$tmp;
unset($tmp);
$sign=-1;
} else $sign = 1;
if ($date1==$date2) return 0;
$days = 0;
$working_days = array(1,2,3,4,5); // Monday-->Friday
$working_hours = array(9, 17.5); // from 9:00 to 17:30 (8.5 hours)
$current_date = $date1;
$beg_h = floor($working_hours[0]);
$beg_m = ($working_hours[0]*60)%60;
$end_h = floor($working_hours[1]);
$end_m = ($working_hours[1]*60)%60;
//In case date1 is on same day of date2
if (mktime(0,0,0,date('n', $date1), date('j', $date1), date('Y', $date1))==mktime(0,0,0,date('n', $date2), date('j', $date2), date('Y', $date2))) {
//If its not working day, then return 0
if (!in_array(date('w', $date1), $working_days)) return 0;
$date0 = mktime($beg_h, $beg_m, 0, date('n', $date1), date('j', $date1), date('Y', $date1));
$date3 = mktime($end_h, $end_m, 0, date('n', $date1), date('j', $date1), date('Y', $date1));
if ($date1<$date0) {
if ($date2<$date0) return 0;
$date1 = $date0;
if ($date2>$date3) $date2=$date3;
return $date2-$date1;
}
if ($date1>$date3) return 0;
if ($date2>$date3) $date2=$date3;
return $date2-$date1;
}
//setup the very next first working time stamp
if (!in_array(date('w',$current_date) , $working_days)) {
// the current day is not a working day
// the current time stamp is set at the beginning of the working day
$current_date = mktime( $beg_h, $beg_m, 0, date('n',$current_date), date('j',$current_date), date('Y',$current_date) );
// search for the next working day
while ( !in_array(date('w',$current_date) , $working_days) ) {
$current_date += 24*3600; // next day
}
} else {
// check if the current timestamp is inside working hours
$date0 = mktime( $beg_h, $beg_m, 0, date('n',$current_date), date('j',$current_date), date('Y',$current_date) );
// it's before working hours, let's update it
if ($current_date<$date0) $current_date = $date0;
$date3 = mktime( $end_h, $end_m, 0, date('n',$current_date), date('j',$current_date), date('Y',$current_date) );
if ($date3<$current_date) {
// outch ! it's after working hours, let's find the next working day
$current_date += 24*3600; // the day after
// and set timestamp as the beginning of the working day
$current_date = mktime( $beg_h, $beg_m, 0, date('n',$current_date), date('j',$current_date), date('Y',$current_date) );
while ( !in_array(date('w',$current_date) , $working_days) ) {
$current_date += 24*3600; // next day
}
}
}
// so, $current_date is now the first working timestamp available...
// calculate the number of seconds from current timestamp to the end of the working day
$date0 = mktime( $end_h, $end_m, 0, date('n',$current_date), date('j',$current_date), date('Y',$current_date) );
$seconds = $date0-$current_date;
// calculate the number of days from the current day to the end day
$date3 = mktime( $beg_h, $beg_m, 0, date('n',$date2), date('j',$date2), date('Y',$date2) );
while ( $current_date < $date3 ) {
if (in_array(date('w',$current_date) , $working_days) ) $days++; // it's a working day
$current_date += 24*3600; // next day
}
$days--; //because we've already count the first day (in $seconds)
// check if end's timestamp is inside working hours
$date0 = mktime( $beg_h, $beg_m, 0, date('n',$date2), date('j',$date2), date('Y',$date2) );
if ((!in_array(date('w', $date2), $working_days)) || ($date2 < $date0)) {
// it's before, so nothing more !
} else {
// is it after ?
$date3 = mktime( $end_h, $end_m, 0, date('n',$date2), date('j',$date2), date('Y',$date2) );
if ($date2>$date3) $date2=$date3;
// calculate the number of seconds from current timestamp to the final timestamp
$tmp = $date2-$date0;
$seconds += $tmp;
}
// calculate the working days in seconds
$seconds += 3600*($working_hours[1]-$working_hours[0])*$days;
return $sign * $seconds;
}
于 2012-12-01T19:00:14.350 回答
1
我知道为时已晚，但我想这个问题还没有解决，我编写了一个代码，可以帮助其他人解决同样的问题。这段代码只是增加工作日，所以，它会跳过午餐时间，周六和周日。
<?php
function recursivoHorarioComercial($data)
{
$hora = date("H", strtotime($data));
if($hora >= 18){
$arrHorarios = array('18'=>'15','19'=>'14','20'=>'13','21'=>'12','22'=>'11','23'=>'10','00'=>'9','01'=>'8','02'=>'7',);
$minutes = (60 * $arrHorarios[$hora]);
$data = date("Y-m-d H:i:s", strtotime("+".$minutes." minutes", strtotime($data)));
}
$diaSemana = date("l", strtotime($data));
if($diaSemana == "Saturday"){
$extraMinutes = (2 * 24 * 60);
$data = date("Y-m-d H:i:s", strtotime("+".$extraMinutes." minutes", strtotime($data)));
} elseif($diaSemana == "Saturday"){
$extraMinutes = (1 * 24 * 60);
$data = date("Y-m-d H:i:s", strtotime("+".$extraMinutes." minutes", strtotime($data)));
}
$horaFim = date("H", strtotime($data));
return $data;
}
function incrementMinutes($data,$minutes)
{
$hours = floor($minutes / 60);
$minutes = ($minutes % 60);
$hid = 8;
$days = floor($hours/$hid);
$horaINi = date("H", strtotime($data));
$hours = $hours - ($days * $hid);
$data = date("Y-m-d H:i:s", strtotime("+".$minutes." minutes", strtotime($data)));
$data = recursivoHorarioComercial($data);
$data = date("Y-m-d H:i:s", strtotime("+".$hours." hours", strtotime($data)));
$data = recursivoHorarioComercial($data);
for ($i = 1; $i <= $days; $i++){
$data = date("Y-m-d H:i:s", strtotime("+1 day", strtotime($data)));
$data = recursivoHorarioComercial($data);
}
$horaFim = date("H", strtotime($data));
if($horaINi <= 12 && $horaFim >= 12 ){
$data = date("Y-m-d H:i:s", strtotime("+60 minutes", strtotime($data)));
}
return $data;
}
$date = "2016-04-06 09:00:00";
$minutos = 24 * 60; //increase 24 hours
echo "Now: ".$data."<br />";
echo "<h2>Modified</h2>";
echo incrementMinutes($date,$minutos)."<br />";
?>
我刚刚测试了这段代码，它工作正常。我希望这可以帮助任何人。
再见
于 2016-04-07T15:39:25.160 回答
0
我知道我参加聚会迟到了，但我试图找到答案并花了几个小时创建自己的答案，我并不是说它完美，如果有人想修改我的代码以使其更清晰，请随意。仍然有一些事情要改变，但它的工作原理。我用碳来协助约会。
function getWorkingHoursDifference($start,$end){
// Declare Business Hours Start
$BusinessHoursStart = '08:30:00';
// Declare Business Hours Start
$BusinessHoursEnd = '17:00:00';
// Set Full Day Span // *Need to Calc from above
$fullDayMinutes = 8.5*60;
// Declare Exclusions
$exclusions = ['Saturday', 'Sunday'];
// Declare Start Time
$TimeStart = Carbon::parse($start);
// Declare End Time
$TimeEnd = Carbon::parse($end);
// Create Start Date
$DateStart = Carbon::parse(date_format($TimeStart, 'Y-m-d'));
// Create End Date
$DateEnd = Carbon::parse(date_format($TimeEnd, 'Y-m-d'));
// Create Period Span
$period = CarbonPeriod::create($DateStart, $DateEnd);
// Create Period Array
$period = $period->toArray();
// Set Day Counter
$dayCount = 0;
// Set Loop Counter
$loopCount = 0;
// Total Minutes
$totalMinutes = 0;
foreach ($period as $date) {$dayCount++;}
// Debugging
//echo '<table class="table"><thead><th>LoopCount</th><th>Period Date</th><th>Day</th><th>Day Start</th><th>Day End</th><th>Minutes</th></thead>';
foreach ($period as $date) {
$dayOnly = date_format($date, 'Y-m-d');
$dayOnlyName = date_format($date, 'l');
$dayStart = Carbon::parse($dayOnly.$BusinessHoursStart);
$dayEnd = Carbon::parse($dayOnly.$BusinessHoursEnd);
// Clean Minutes
$minutes = 0;
// Loop Counter
$loopCount++;
// Check Excluded Days
if(in_array($dayOnlyName, $exclusions)){$minutes = 0;}
// Check if its a single day
elseif($loopCount == 1 && $dayCount == 1){
if($TimeStart < $dayStart && $TimeEnd > $dayStart && $TimeEnd < $dayEnd){$minutes = $dayStart->diffInMinutes($TimeEnd);}
// If the time start is after business hours then 0
elseif($TimeStart > $dayEnd){$minutes = 0;}
// If the time start is after start of the day, but before closing hours
elseif($TimeStart > $dayStart){$minutes = $TimeStart->diffInMinutes($TimeEnd);}
}
// First Day
elseif($loopCount == 1){
// If the Time Start is before business hours, count as a full day.
if($TimeStart < $dayStart){$minutes = $fullDayMinutes;}
// If the time start is after business hours then 0
elseif($TimeStart > $dayEnd){$minutes = 0;}
// If the time start is after start of the day, but before closing hours
elseif($TimeStart > $dayStart){$minutes = $TimeStart->diffInMinutes($dayEnd);}
}
// Last Day (loopCounter Matches Total Day Counter)
elseif($loopCount == $dayCount){
// If the Time is before the start of the business day
if($TimeEnd< $dayStart){$minutes = 0;}
// If the End Time is after the day end then full day is counted
elseif($TimeEnd > $dayEnd){$minutes = $fullDayMinutes;}
// If the End time is after the start but before the end of business hours, work out the difference
elseif($TimeEnd > $dayStart){$minutes = $dayStart->diffInMinutes($TimeEnd);}
}
// All other days in the middle
else{$minutes = $fullDayMinutes;}
// Debugging
//echo '<tr><td>'.$loopCount.'</td><td>'.$date.'</td><td>'.$dayOnlyName.'</td><td>'.$dayStart.'</td><td>'.$dayEnd.'</td><td>'.$minutes.'</td></tr>';
$totalMinutes = $totalMinutes + $minutes;
}
//echo '</table>';
return $totalMinutes;
}
于 2019-08-15T08:58:16.890 回答