如果你不喜欢公认的答案或流行的答案，那么试试这个

function secondsToTime($seconds_time)
{
    if ($seconds_time < 24 * 60 * 60) {
        return gmdate('H:i:s', $seconds_time);
    } else {
        $hours = floor($seconds_time / 3600);
        $minutes = floor(($seconds_time - $hours * 3600) / 60);
        $seconds = floor($seconds_time - ($hours * 3600) - ($minutes * 60));
        return "$hours:$minutes:$seconds";
    }
}

secondsToTime(108620); // 30:10:20

使用DateTime的一个简单方法是:

    $time = 60; //sec.
    $now = time();
    $rep = new DateTime('@'.$now);
    $diff = new DateTime('@'.($now+$time));
    $return = $diff->diff($rep)->format($format);

    //output:  01:04:65

这是一个简单的解决方案，让您能够使用DateTime的格式方法。

任何人在未来寻找这个，这给了最初的海报要求的格式。

$init = 685;
$hours = floor($init / 3600);
$hrlength=strlen($hours);
if ($hrlength==1) {$hrs="0".$hours;}
else {$hrs=$hours;} 

$minutes = floor(($init / 60) % 60);
$minlength=strlen($minutes);
if ($minlength==1) {$mins="0".$minutes;}
else {$mins=$minutes;} 

$seconds = $init % 60;
$seclength=strlen($seconds);
if ($seclength==1) {$secs="0".$seconds;}
else {$secs=$seconds;} 

echo "$hrs:$mins:$secs";

这个函数是有用的，你可以扩展它:

function formatSeconds($seconds) {

if(!is_integer($seconds)) {
    return FALSE;
}

$fmt = "";

$days = floor($seconds / 86400);
if($days) {
    $fmt .= $days."D ";
    $seconds %= 86400;
}

$hours = floor($seconds / 3600);
if($hours) {
    $fmt .= str_pad($hours, 2, '0', STR_PAD_LEFT).":";
    $seconds %= 3600;
}

$mins = floor($seconds / 60 );
if($mins) {
    $fmt .= str_pad($mins, 2, '0', STR_PAD_LEFT).":";
    $seconds %= 60;
}

$fmt .= str_pad($seconds, 2, '0', STR_PAD_LEFT);

return $fmt;}

解决方案来自:https://gist.github.com/SteveJobzniak/c91a8e2426bac5cb9b0cbc1bdbc45e4b

这段代码尽可能地避免了乏味的函数调用和逐条的字符串构建，以及人们为此而创建的大而笨重的函数。

它返回格式为“1h05m00s”的输出，并使用前导零表示分和秒，只要在它们之前有另一个非零时间组件。

它还跳过所有空的前导组件，以避免给你无用的信息，如“0h00m01s”(相反，它将显示为“1s”)。

示例结果：“1s”、“1m00s”、“19m08s”、“1h00m00s”、“4h08m39s”。

$duration = 1; // values 0 and higher are supported!
$converted = [
    'hours' => floor( $duration / 3600 ),
    'minutes' => floor( ( $duration / 60 ) % 60 ),
    'seconds' => ( $duration % 60 )
];
$result = ltrim( sprintf( '%02dh%02dm%02ds', $converted['hours'], $converted['minutes'], $converted['seconds'] ), '0hm' );
if( $result == 's' ) { $result = '0s'; }

如果你想让代码更短(但可读性更差)，你可以避免$转换数组，而是直接将值放在sprintf()调用中，如下所示:

$duration = 1; // values 0 and higher are supported!
$result = ltrim( sprintf( '%02dh%02dm%02ds', floor( $duration / 3600 ), floor( ( $duration / 60 ) % 60 ), ( $duration % 60 ) ), '0hm' );
if( $result == 's' ) { $result = '0s'; }

在上面的两个代码段中，持续时间必须为0或更高。不支持负持续时间。但是你可以使用下面的代码来处理负持续时间:

$duration = -493; // negative values are supported!
$wasNegative = FALSE;
if( $duration < 0 ) { $wasNegative = TRUE; $duration = abs( $duration ); }
$converted = [
    'hours' => floor( $duration / 3600 ),
    'minutes' => floor( ( $duration / 60 ) % 60 ),
    'seconds' => ( $duration % 60 )
];
$result = ltrim( sprintf( '%02dh%02dm%02ds', $converted['hours'], $converted['minutes'], $converted['seconds'] ), '0hm' );
if( $result == 's' ) { $result = '0s'; }
if( $wasNegative ) { $result = "-{$result}"; }
// $result is now "-8m13s"

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 )