我要做的是将跨越24小时的小时/分钟/秒相加,例如:
12:39:25
08:22:10
11:08:50
07:33:05
我希望它返回的是"39:43:30",而不是1970年的日期。下面是我目前正在使用的代码(注意,它来自一个类,而不仅仅是一个函数)。
private function add_time($time1, $time2)
{
$first_exploded = explode(":", $time1);
$second_exploded = explode(":", $time2);
$first_stamp = mktime($first_exploded[0],$first_exploded[1],$first_exploded[2],1,1,1970);
$second_stamp = mktime($second_exploded[0],$second_exploded[1],$second_exploded[2],1,1,1970);
$time_added = $first_stamp + $second_stamp;
$sum_time = date("H:i:s",$time_added);
return $sum_time;
}
如有任何建议,不胜感激。
日期函数总是围绕天/月/年工作。你想要的是一个简单的数学函数,没有测试它,但应该把它弄清楚。
private function add_time($base, $toadd) {
$base = explode(':', $base);
$toadd = explode(':', $toadd);
$res = array();
$res[0] = $base[0] + $toadd[0];
$res[1] = $base[1] + $toadd[1];
$res[2] = $base[2] + $toadd[2];
// Seconds
while($res[2] >= 60) {
$res[1] += 1;
$res[2] -= 60;
}
// Minutes
while($res[1] >= 60) {
$res[0] += 1;
$res[1] -= 60;
}
return implode(':', $res);
}
这里有一个很好的小函数,可以将数组中传递的任何次数相加:-
function addTimes(Array $times)
{
$total = 0;
foreach($times as $time){
list($hours, $minutes, $seconds) = explode(':', $time);
$hour = (int)$hours + ((int)$minutes/60) + ((int)$seconds/3600);
$total += $hour;
}
$h = floor($total);
$total -= $h;
$m = floor($total * 60);
$total -= $m/60;
$s = floor($total * 3600);
return "$h:$m:$s";
}
这样使用:-
$times = array('12:39:25', '08:22:10', '11:08:50', '07:33:05',);
var_dump(addTimes($times));
输出:-
string '39:43:30' (length=8)