gmtdate()函数没有为我工作，因为我在跟踪一个项目上的工作时间，如果超过24小时，你会得到24小时后减去的剩余量。换句话说，37小时变成了13小时。(如上所述，Glavic -谢谢你的例子!) 这个方法很有效:

Convert seconds to format by 'foot' no limit :
$seconds = 8525;
$H = floor($seconds / 3600);
$i = ($seconds / 60) % 60;
$s = $seconds % 60;
echo sprintf("%02d:%02d:%02d", $H, $i, $s);
# 02:22:05

// TEST
// 1 Day 6 Hours 50 Minutes 31 Seconds ~ 111031 seconds

$time = 111031; // time duration in seconds

$days = floor($time / (60 * 60 * 24));
$time -= $days * (60 * 60 * 24);

$hours = floor($time / (60 * 60));
$time -= $hours * (60 * 60);

$minutes = floor($time / 60);
$time -= $minutes * 60;

$seconds = floor($time);
$time -= $seconds;

echo "{$days}d {$hours}h {$minutes}m {$seconds}s"; // 1d 6h 50m 31s

下面是一个处理负秒和超过1天的秒的一行程序。

sprintf("%s:%'02s:%'02s\n", intval($seconds/60/60), abs(intval(($seconds%3600) / 60)), abs($seconds%60));

例如:

$seconds= -24*60*60 - 2*60*60 - 3*60 - 4; // minus 1 day 2 hours 3 minutes 4 seconds
echo sprintf("%s:%'02s:%'02s\n", intval($seconds/60/60), abs(intval(($seconds%3600) / 60)), abs($seconds%60));

输出:26:03:04

以防其他人正在寻找一个简单的函数来返回格式化良好的格式(我知道这不是OP要求的格式)，这就是我刚刚想到的。感谢@mughal的代码，这是基于。

function format_timer_result($time_in_seconds){
    $time_in_seconds = ceil($time_in_seconds);

    // Check for 0
    if ($time_in_seconds == 0){
        return 'Less than a second';
    }

    // Days
    $days = floor($time_in_seconds / (60 * 60 * 24));
    $time_in_seconds -= $days * (60 * 60 * 24);

    // Hours
    $hours = floor($time_in_seconds / (60 * 60));
    $time_in_seconds -= $hours * (60 * 60);

    // Minutes
    $minutes = floor($time_in_seconds / 60);
    $time_in_seconds -= $minutes * 60;

    // Seconds
    $seconds = floor($time_in_seconds);

    // Format for return
    $return = '';
    if ($days > 0){
        $return .= $days . ' day' . ($days == 1 ? '' : 's'). ' ';
    }
    if ($hours > 0){
        $return .= $hours . ' hour' . ($hours == 1 ? '' : 's') . ' ';
    }
    if ($minutes > 0){
        $return .= $minutes . ' minute' . ($minutes == 1 ? '' : 's') . ' ';
    }
    if ($seconds > 0){
        $return .= $seconds . ' second' . ($seconds == 1 ? '' : 's') . ' ';
    }
    $return = trim($return);

    return $return;
}

不知道为什么还没有提出这个建议，但是这里有一个使用DateInterval的变体，它的优点是秒> 86400不是问题(只是不要期望小时< 24)

noteable:该函数返回DateInterval对象，因此输出格式由调用者决定

function seconds2interval($seconds) {
    $s = intval($seconds);
    $hrs = floor($s / 3600);
    $min = floor(($s - ($hrs * 3600)) / 60);
    $sec = $s % 60;
    
    $duration = sprintf('PT%dH%dM%dS', $hrs, $min, $sec);
    $d = new DateInterval($duration);
    
    return $d;
}

$d = seconds2interval(93837.3113);  
echo $d->format('%H:%I:%s'); // 26:03:57

写这样的函数返回一个数组

function secondsToTime($seconds) {

  // extract hours
  $hours = floor($seconds / (60 * 60));

  // extract minutes
  $divisor_for_minutes = $seconds % (60 * 60);
  $minutes = floor($divisor_for_minutes / 60);

  // extract the remaining seconds
  $divisor_for_seconds = $divisor_for_minutes % 60;
  $seconds = ceil($divisor_for_seconds);

  // return the final array
  $obj = array(
      "h" => (int) $hours,
      "m" => (int) $minutes,
      "s" => (int) $seconds,
   );

  return $obj;
}

然后像这样简单地调用函数:

secondsToTime(100);

输出是

Array ( [h] => 0 [m] => 1 [s] => 40 )