以秒为单位计算时间差,不包括周末和PHP中的时间段
我已经放弃了在MySQL中这样做的尝试,取而代之的是,我将把结果注入到我的结果数组中,这将在稍后的JS应用程序中使用 我有一个循环来遍历每个结果:以秒为单位计算时间差,不包括周末和PHP中的时间段,php,mysql,date,datetime,Php,Mysql,Date,Datetime,我已经放弃了在MySQL中这样做的尝试,取而代之的是,我将把结果注入到我的结果数组中,这将在稍后的JS应用程序中使用 我有一个循环来遍历每个结果: $data = $result->fetchAll(); foreach($data as $i => $row) { // Something $data[$i]['Age'] = $row['Age']; } 我希望它将$data[$I]['Age']这是一个日期时间和当前日
$data = $result->fetchAll();
foreach($data as $i => $row) {
// Something
$data[$i]['Age'] = $row['Age'];
}
我希望它将$data[$I]['Age']这是一个日期时间和当前日期时间之间的秒数相加
通常这很简单,但如何排除周末和16:30到07:30之间的时间?我不认为这太具体或无趣,因此以下是我的一位同事给出的答案:
public static function calculateTime($startDate, $endDate) {//Returns Seconds Passed
date_default_timezone_set('Europe/London');
//Opening Hours - Can pull from database depending on users working hours
$workingHoursOpen = new DateTime('07:30:00');
$workingHoursClose = new DateTime('16:30:00');
//Time worked from and to
$timeStarted = strtotime(date('H:i:s', strtotime($endDate)));
$timeFinished = strtotime(date('H:i:s', strtotime($startDate)));
$workingSeconds = $workingHoursClose->getTimestamp() - $workingHoursOpen->getTimestamp();
$workingSecondsv2 = $timeFinished - $timeStarted;
//Option to send array of holidays (3rd param)
$workingDays = Util::getWorkingDays(date('Y-m-d', strtotime($startDate)), date('Y-m-d', strtotime($endDate)), array());
$totalWorkingSeconds = $workingDays * $workingSeconds; //Working days * 9 hours
$secondsClosed = 0;
$i = 0;
while ($i < $workingDays) {
$secondsClosed = $secondsClosed - (15 * 3600);
$i++;
}
$secondsPassed = ($workingSecondsv2 - $totalWorkingSeconds) + (9 * 3600);
$secondsPassed = -1 * ($secondsPassed); // Invert number (was giving -XX)
return $secondsPassed;
}
public static function getWorkingDays($startDate, $endDate, $holidays) {
// do strtotime calculations just once
$endDate = strtotime($endDate);
$startDate = strtotime($startDate);
$days = ($endDate - $startDate) / 86400 + 1;
$no_full_weeks = floor($days / 7);
$no_remaining_days = fmod($days, 7);
//It will return 1 if it's Monday,.. ,7 for Sunday
$the_first_day_of_week = date("N", $startDate);
$the_last_day_of_week = date("N", $endDate);
//---->The two can be equal in leap years when february has 29 days, the equal sign is added here
//In the first case the whole interval is within a week, in the second case the interval falls in two weeks.
if ($the_first_day_of_week <= $the_last_day_of_week) {
if ($the_first_day_of_week <= 6 && 6 <= $the_last_day_of_week)
$no_remaining_days--;
if ($the_first_day_of_week <= 7 && 7 <= $the_last_day_of_week)
$no_remaining_days--;
}
else {
// (edit by Tokes to fix an edge case where the start day was a Sunday
// and the end day was NOT a Saturday)
// the day of the week for start is later than the day of the week for end
if ($the_first_day_of_week == 7) {
// if the start date is a Sunday, then we definitely subtract 1 day
$no_remaining_days--;
if ($the_last_day_of_week == 6) {
// if the end date is a Saturday, then we subtract another day
$no_remaining_days--;
}
} else {
// the start date was a Saturday (or earlier), and the end date was (Mon..Fri)
// so we skip an entire weekend and subtract 2 days
$no_remaining_days -= 2;
}
}
//The no. of business days is: (number of weeks between the two dates) * (5 working days) + the remainder
//---->february in none leap years gave a remainder of 0 but still calculated weekends between first and last day, this is one way to fix it
$workingDays = $no_full_weeks * 5;
if ($no_remaining_days > 0) {
$workingDays += $no_remaining_days;
}
//We subtract the holidays
foreach ($holidays as $holiday) {
$time_stamp = strtotime($holiday);
//If the holiday doesn't fall in weekend
if ($startDate <= $time_stamp && $time_stamp <= $endDate && date("N", $time_stamp) != 6 && date("N", $time_stamp) != 7)
$workingDays--;
}
return $workingDays;
}
100%工作,可以扩展,从数据库中输入工作时间。假期如何?在哪个日历下?可能的重复应该是一个很好的起点。您可以进行标准的加法,然后仔细计算间隔期间折扣时间跨度的持续时间,并从总和中减去该时间跨度。@imperium2335:否。此外,verbumSapienti提供了一个有用的起点,而不是解决方案。在我看来,极不可能有人会费心为一个非常具体、非常无趣的问题编写解决方案。