我有一个php函数来计算时间戳（开始日期）到当前（结束）日期，它将返回1小时30分2秒格式。我想要实现的是只计算一个工作日的上午 8 点到下午 5 点。任何超出的内容都不会被计算在内。这是我拥有的 php 代码。
class duration_computation {
function duration( $time ) {
$d[0] = array(1, "s");
$d[1] = array(60, "min");
$d[2] = array(3600, "hr");
$d[3] = array(86400, "dy");
$d[4] = array(604800, "wk");
$d[5] = array(2592000, "mth");
$d[6] = array(31104000, "yr");
$numbers = array();
$result = "";
$now = time();
$time_difference = ( $now - $time );
$seconds_left = $time_difference;
for ( $i = 6; $i > -1; $i-- ) {
$numbers[$i] = intval( $seconds_left / $d[$i][0] );
$seconds_left -= ( $numbers[$i] * $d[$i][0] );
if ( $numbers[$i] != 0 ) {
$result.= abs($numbers[$i]) . "" . $d[$i][1] . (($numbers[$i]>1)?'':'') ." ";
}
}
return $result;
}
}
$duration = new duration_computation();
echo $duration->duration($trail->duration);
忘记date(), strtotime(), time(), etc.函数，使用 DateTime ：
使用示例 ：
$from = '2013-09-06 15:45:32';
$to = '2013-09-14 21:00:00';
echo some_func_name($from, $to);
输出：
1 day, 22 hours, 14 minutes, 28 seconds
功能：
function some_func_name($from, $to) {
$workingDays = [1, 2, 3, 4, 5]; # date format = N
$workingHours = ['from' => ['08', '00'], 'to' => ['17', '00']];
$start = new DateTime($from);
$end = new DateTime($to);
$startP = clone $start;
$startP->setTime(0, 0, 0);
$endP = clone $end;
$endP->setTime(23, 59, 59);
$interval = new DateInterval('P1D');
$periods = new DatePeriod($startP, $interval, $endP);
$sum = [];
foreach ($periods as $i => $period) {
if (!in_array($period->format('N'), $workingDays)) continue;
$startT = clone $period;
$startT->setTime($workingHours['from'][0], $workingHours['from'][1]);
if (!$i && $start->diff($startT)->invert) $startT = $start;
$endT = clone $period;
$endT->setTime($workingHours['to'][0], $workingHours['to'][1]);
if (!$end->diff($endT)->invert) $endT = $end;
#echo $startT->format('Y-m-d H:i') . ' - ' . $endT->format('Y-m-d H:i') . "'n"; # debug
$diff = $startT->diff($endT);
if ($diff->invert) continue;
foreach ($diff as $k => $v) {
if (!isset($sum[$k])) $sum[$k] = 0;
$sum[$k] += $v;
}
}
if (!$sum) return 'ccc, no time on job?';
$spec = "P{$sum['y']}Y{$sum['m']}M{$sum['d']}DT{$sum['h']}H{$sum['i']}M{$sum['s']}S";
$interval = new DateInterval($spec);
$startS = new DateTime;
$endS = clone $startS;
$endS->sub($interval);
$diff = $endS->diff($startS);
$labels = [
'y' => 'year',
'm' => 'month',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
];
$return = [];
foreach ($labels as $k => $v) {
if ($diff->$k) {
$return[] = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
}
}
return implode(', ', $return);
}
这个功能可以更短/更好;但这是你现在的工作;)