如何平均负值PHP的时间？

Question

我有一个数组说：

array("10:10:20", "-20:00:10", "14:15:00", "08:10:00"); 

我尝试使用strtime，但它给我的错误，因为它不能负的值转换为时间戳。有没有其他办法？

Answer 1

<?php 
$moredownvotes = array("10:10:20", "-20:00:10", "14:15:00", "08:10:00"); 

$nb_lol = count($moredownvotes); 

for ($i = 0; $i < ($nb_lol - 1); ++$i) 
{ 
    $pattern = '/^-(.*)/'; 
    $replacement = '${1}'; 
    $moredownvotes[$i] = preg_replace($pattern, $replacement, $moredownvotes[$i]); 
} 

print "<pre>"; 
print_r($moredownvotes); 
print "</pre>"; 
?> 

结果：

Array 
(
    [0] => 10:10:20 
    [1] => 20:00:10 
    [2] => 14:15:00 
    [3] => 08:10:00 
) 

NOW u可以使用日期时间函数或W/Eü要

$date = DateTime::createFromFormat('H:i:s', $moredownvotes[$i]); 
echo date_format($date, 'H:i:s'); 

** **结果：10:10:20

PHP.net createFromFormat function

Answer 2

重读这个问题后，我做了一个小片段，平均值是一个数字的总和。所以我猜这应该是正确的。要绕过负面的时间戳，您只需计算绝对值，使其为负值并将其添加到总和中。

<?php 

    $timings = array("10:10:20", "-20:00:10", "14:15:00", "08:10:00"); 
    //$timings = array("-10:10:20", "-10:10:20", "-10:10:20", "-10:10:20"); 

    $totalTime = 0; 
    foreach($timings as $timing) { 
     $isNegativeTime = strpos($timing, '-') !== false; 
     $time = strtotime(str_replace('-', '', $timing)); 
     if ($isNegativeTime) $time *= -1; 
     $totalTime += $time; 
    } 

    $avgTiming = $totalTime/count($timings); 
    $isNegativeAvgTime = $avgTiming < 0; 
    $avgTiming = date('h:i:s', abs($avgTiming)); 
    echo ($isNegativeAvgTime ? '-':'').$avgTiming;