你可以使用gmdate()函数:

echo gmdate("H:i:s", 685);

仅当秒数小于86400(1天)时使用函数gmdate():

$seconds = 8525;
echo gmdate('H:i:s', $seconds);
# 02:22:05

参见： gmdate（）

运行Demo

将秒转换为“脚”格式无限制*:

$seconds = 8525;
$H = floor($seconds / 3600);
$i = ($seconds / 60) % 60;
$s = $seconds % 60;
echo sprintf("%02d:%02d:%02d", $H, $i, $s);
# 02:22:05

参见:floor()， sprintf()，算术运算符

运行Demo

DateTime扩展的示例使用:

$seconds = 8525;
$zero    = new DateTime("@0");
$offset  = new DateTime("@$seconds");
$diff    = $zero->diff($offset);
echo sprintf("%02d:%02d:%02d", $diff->days * 24 + $diff->h, $diff->i, $diff->s);
# 02:22:05

参见:DateTime::__construct()， DateTime::modify()， clone， sprintf ()

运行Demo

MySQL示例的结果范围被限制为TIME数据类型，从-838:59:59到838:59:59:

SELECT SEC_TO_TIME(8525);
# 02:22:05

看:SEC_TO_TIME

运行Demo

PostgreSQL的例子:

SELECT TO_CHAR('8525 second'::interval, 'HH24:MI:SS');
# 02:22:05

运行Demo

See:

    /** 
     * Convert number of seconds into hours, minutes and seconds 
     * and return an array containing those values 
     * 
     * @param integer $inputSeconds Number of seconds to parse 
     * @return array 
     */ 

    function secondsToTime($inputSeconds) {

        $secondsInAMinute = 60;
        $secondsInAnHour  = 60 * $secondsInAMinute;
        $secondsInADay    = 24 * $secondsInAnHour;

        // extract days
        $days = floor($inputSeconds / $secondsInADay);

        // extract hours
        $hourSeconds = $inputSeconds % $secondsInADay;
        $hours = floor($hourSeconds / $secondsInAnHour);

        // extract minutes
        $minuteSeconds = $hourSeconds % $secondsInAnHour;
        $minutes = floor($minuteSeconds / $secondsInAMinute);

        // extract the remaining seconds
        $remainingSeconds = $minuteSeconds % $secondsInAMinute;
        $seconds = ceil($remainingSeconds);

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

From:将秒转换为日、时、分和秒

在java中你可以使用这种方法。

   private String getHmaa(long seconds) {
    String string;
    int hours = (int) seconds / 3600;
    int remainder = (int) seconds - hours * 3600;
    int mins = remainder / 60;
    //remainder = remainder - mins * 60;
    //int secs = remainder;

    if (hours < 12 && hours > 0) {
        if (mins < 10) {
            string = String.valueOf((hours < 10 ? "0" + hours : hours) + ":" + (mins > 0 ? "0" + mins : "0") + " AM");
        } else {
            string = String.valueOf((hours < 10 ? "0" + hours : hours) + ":" + (mins > 0 ? mins : "0") + " AM");
        }
    } else if (hours >= 12) {
        if (mins < 10) {
            string = String.valueOf(((hours - 12) < 10 ? "0" + (hours - 12) : ((hours - 12) == 12 ? "0" : (hours - 12))) + ":" + (mins > 0 ? "0" + mins : "0") + ((hours - 12) == 12 ? " AM" : " PM"));
        } else {
            string = String.valueOf(((hours - 12) < 10 ? "0" + (hours - 12) : ((hours - 12) == 12 ? "0" : (hours - 12))) + ":" + (mins > 0 ? mins : "0") + ((hours - 12) == 12 ? " AM" : " PM"));
        }
    } else {
        if (mins < 10) {
            string = String.valueOf("0" + ":" + (mins > 0 ? "0" + mins : "0") + " AM");
        } else {
            string = String.valueOf("0" + ":" + (mins > 0 ? mins : "0") + " AM");
        }
    }
    return string;
}

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"

如果你想创建一个音频/视频持续时间字符串，如YouTube等，你可以这样做:

($seconds >= 60) ?ltrim(函数(“我:年代”,美元秒),“0”):函数(“0:年代”,美元秒)

将返回如下字符串:

55.55 => '0:55'
100   => '1:40'

在>= 24小时内可能不会很好地工作。