我想将变量$uptime（秒）转换为天，小时，分钟和秒。
示例：
$uptime = 1640467;
结果应该是：
18 days 23 hours 41 minutes
答案 0 :(得分：198)
这可以通过DateTime class
使用：
echo secondsToTime(1640467);
# 18 days, 23 hours, 41 minutes and 7 seconds
<强>功能：
function secondsToTime($seconds) {
$dtF = new \DateTime('@0');
$dtT = new \DateTime("@$seconds");
return $dtF->diff($dtT)->format('%a days, %h hours, %i minutes and %s seconds');
}
答案 1 :(得分：41)
这是重写的功能，包括天。我还更改了变量名，使代码更容易理解......
/**
* 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 $obj;
}
来源：CodeAid（） - http://codeaid.net/php/convert-seconds-to-hours-minutes-and-seconds-(php)
答案 2 :(得分：18)
根据Julian Moreno的回答，但改为将响应作为字符串（不是数组）给出，只包括所需的时间间隔而不是假设复数。
这个和最高投票答案的区别在于：
259264秒，此代码会给出
3天，1分钟，4秒
在259264秒内，最高投票回答（由Glavić提供）会给出
3天， 0小时，1分钟 s 4秒
function secondsToTime($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);
// Format and return
$timeParts = [];
$sections = [
'day' => (int)$days,
'hour' => (int)$hours,
'minute' => (int)$minutes,
'second' => (int)$seconds,
];
foreach ($sections as $name => $value){
if ($value > 0){
$timeParts[] = $value. ' '.$name.($value == 1 ? '' : 's');
}
}
return implode(', ', $timeParts);
}
我希望这有助于某人。
答案 3 :(得分：11)
这是一个简单的8行PHP函数，它将一些秒转换为人类可读的字符串，包括大量秒的月数：
答案 4 :(得分：9)
gmdate("d H:i:s",1640467);
结果将是19 23:41:07。当它比正常日仅一秒钟时，它会增加1天的日值。这就是为什么它显示19.你可以根据需要分解结果并修复它。
答案 5 :(得分：6)
这里有一些很好的答案，但没有一个满足我的需求。我在Glavic's answer上构建了一些我需要的额外功能;
您可以看到代码的运行版本 here 。
function secondsToHumanReadable(int $seconds, int $requiredParts = null)
{
$from = new \DateTime('@0');
$to = new \DateTime("@$seconds");
$interval = $from->diff($to);
$str = '';
$parts = [
'y' => 'year',
'm' => 'month',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
];
$includedParts = 0;
foreach ($parts as $key => $text) {
if ($requiredParts && $includedParts >= $requiredParts) {
break;
}
$currentPart = $interval->{$key};
if (empty($currentPart)) {
continue;
}
if (!empty($str)) {
$str .= ', ';
}
$str .= sprintf('%d %s', $currentPart, $text);
if ($currentPart > 1) {
// handle plural
$str .= 's';
}
$includedParts++;
}
return $str;
}
答案 6 :(得分：6)
最简单的方法是创建一个方法，从当前时间$ now返回DateTime :: diff相对时间的DateInterval，该时间为$ seconds，然后可以链接和格式化。例如： -
public function toDateInterval($seconds) {
return date_create('@' . (($now = time()) + $seconds))->diff(date_create('@' . $now));
}
现在将您的方法调用链接到DateInterval :: format
echo $this->toDateInterval(1640467)->format('%a days %h hours %i minutes'));
结果：
18 days 23 hours 41 minutes
答案 7 :(得分：5)
简短，简单，可靠：
function secondsToDHMS($seconds) {
$s = (int)$seconds;
return sprintf('%d:%02d:%02d:%02d', $s/86400, $s/3600%24, $s/60%60, $s%60);
}
答案 8 :(得分：3)
Laravel 示例
Carbon 支持 700 多个语言环境
\Carbon\CarbonInterval::seconds(1640467)->cascade()->forHumans(); //2 weeks 4 days 23 hours 41 minutes 7 seconds
答案 9 :(得分：3)
function seconds_to_time($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);
//create string HH:MM:SS
$ret = $hours.":".$minutes.":".$seconds;
return($ret);
}
答案 10 :(得分：3)
虽然这是一个非常古老的问题 - 人们可能会发现这些有用（不是写得很快）：
function d_h_m_s__string1($seconds)
{
$ret = '';
$divs = array(86400, 3600, 60, 1);
for ($d = 0; $d < 4; $d++)
{
$q = (int)($seconds / $divs[$d]);
$r = $seconds % $divs[$d];
$ret .= sprintf("%d%s", $q, substr('dhms', $d, 1));
$seconds = $r;
}
return $ret;
}
function d_h_m_s__string2($seconds)
{
if ($seconds == 0) return '0s';
$can_print = false; // to skip 0d, 0d0m ....
$ret = '';
$divs = array(86400, 3600, 60, 1);
for ($d = 0; $d < 4; $d++)
{
$q = (int)($seconds / $divs[$d]);
$r = $seconds % $divs[$d];
if ($q != 0) $can_print = true;
if ($can_print) $ret .= sprintf("%d%s", $q, substr('dhms', $d, 1));
$seconds = $r;
}
return $ret;
}
function d_h_m_s__array($seconds)
{
$ret = array();
$divs = array(86400, 3600, 60, 1);
for ($d = 0; $d < 4; $d++)
{
$q = $seconds / $divs[$d];
$r = $seconds % $divs[$d];
$ret[substr('dhms', $d, 1)] = $q;
$seconds = $r;
}
return $ret;
}
echo d_h_m_s__string1(0*86400+21*3600+57*60+13) . "\n";
echo d_h_m_s__string2(0*86400+21*3600+57*60+13) . "\n";
$ret = d_h_m_s__array(9*86400+21*3600+57*60+13);
printf("%dd%dh%dm%ds\n", $ret['d'], $ret['h'], $ret['m'], $ret['s']);
结果：
0d21h57m13s
21h57m13s
9d21h57m13s
答案 11 :(得分：2)
应该排除0值并设置正确的单数/复数值的解决方案
use DateInterval;
use DateTime;
class TimeIntervalFormatter
{
public static function fromSeconds($seconds)
{
$seconds = (int)$seconds;
$dateTime = new DateTime();
$dateTime->sub(new DateInterval("PT{$seconds}S"));
$interval = (new DateTime())->diff($dateTime);
$pieces = explode(' ', $interval->format('%y %m %d %h %i %s'));
$intervals = ['year', 'month', 'day', 'hour', 'minute', 'second'];
$result = [];
foreach ($pieces as $i => $value) {
if (!$value) {
continue;
}
$periodName = $intervals[$i];
if ($value > 1) {
$periodName .= 's';
}
$result[] = "{$value} {$periodName}";
}
return implode(', ', $result);
}
}
答案 12 :(得分：2)
function convert($seconds){
$string = "";
$days = intval(intval($seconds) / (3600*24));
$hours = (intval($seconds) / 3600) % 24;
$minutes = (intval($seconds) / 60) % 60;
$seconds = (intval($seconds)) % 60;
if($days> 0){
$string .= "$days days ";
}
if($hours > 0){
$string .= "$hours hours ";
}
if($minutes > 0){
$string .= "$minutes minutes ";
}
if ($seconds > 0){
$string .= "$seconds seconds";
}
return $string;
}
echo convert(3744000);
答案 13 :(得分：1)
Glavić's excellent solution的扩展版本，具有整数验证，解决了1 s问题，以及数年和数月的额外支持，代价是减少了计算机解析的友好性，有利于更加人性化：
<?php
function secondsToHumanReadable(/*int*/ $seconds)/*: string*/ {
//if you dont need php5 support, just remove the is_int check and make the input argument type int.
if(!\is_int($seconds)){
throw new \InvalidArgumentException('Argument 1 passed to secondsToHumanReadable() must be of the type int, '.\gettype($seconds).' given');
}
$dtF = new \DateTime ( '@0' );
$dtT = new \DateTime ( "@$seconds" );
$ret = '';
if ($seconds === 0) {
// special case
return '0 seconds';
}
$diff = $dtF->diff ( $dtT );
foreach ( array (
'y' => 'year',
'm' => 'month',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second'
) as $time => $timename ) {
if ($diff->$time !== 0) {
$ret .= $diff->$time . ' ' . $timename;
if ($diff->$time !== 1 && $diff->$time !== -1 ) {
$ret .= 's';
}
$ret .= ' ';
}
}
return substr ( $ret, 0, - 1 );
}
var_dump(secondsToHumanReadable(1*60*60*2+1)); - ＆gt; string(16) "2 hours 1 second"
答案 14 :(得分：1)
我不知道为什么这些答案有些冗长或复杂。这是使用DateTime Class的一个。有点类似于radzserg的答案。这只会显示必要的单位，负数时将带有“ ago”后缀...
function calctime($seconds = 0) {
$datetime1 = date_create("@0");
$datetime2 = date_create("@$seconds");
$interval = date_diff($datetime1, $datetime2);
if ( $interval->y >= 1 ) $thetime[] = pluralize( $interval->y, 'year' );
if ( $interval->m >= 1 ) $thetime[] = pluralize( $interval->m, 'month' );
if ( $interval->d >= 1 ) $thetime[] = pluralize( $interval->d, 'day' );
if ( $interval->h >= 1 ) $thetime[] = pluralize( $interval->h, 'hour' );
if ( $interval->i >= 1 ) $thetime[] = pluralize( $interval->i, 'minute' );
if ( $interval->s >= 1 ) $thetime[] = pluralize( $interval->s, 'second' );
return isset($thetime) ? implode(' ', $thetime) . ($interval->invert ? ' ago' : '') : NULL;
}
function pluralize($count, $text) {
return $count . ($count == 1 ? " $text" : " ${text}s");
}
// Examples:
// -86400 = 1 day ago
// 12345 = 3 hours 25 minutes 45 seconds
// 987654321 = 31 years 3 months 18 days 4 hours 25 minutes 21 seconds
编辑：如果您想将上面的示例压缩为使用更少的变量/空间（以牺牲易读性为代价），这是一个替代版本，其功能相同：
function calctime($seconds = 0) {
$interval = date_diff(date_create("@0"),date_create("@$seconds"));
foreach (array('y'=>'year','m'=>'month','d'=>'day','h'=>'hour','i'=>'minute','s'=>'second') as $format=>$desc) {
if ($interval->$format >= 1) $thetime[] = $interval->$format . ($interval->$format == 1 ? " $desc" : " {$desc}s");
}
return isset($thetime) ? implode(' ', $thetime) . ($interval->invert ? ' ago' : '') : NULL;
}
答案 15 :(得分：0)
添加了一些格式修改自 Glavić 对 Facebook 风格的帖子计数时间的出色回答....
function secondsToTime($seconds) {
$dtF = new \DateTime('@0');
$dtT = new \DateTime("@$seconds");
switch($seconds){
case ($seconds<60*60*24): // if time is less than one day
return $dtF->diff($dtT)->format('%h hours, %i minutes, %s seconds');
break;
case ($seconds<60*60*24*31 && $seconds>60*60*24): // if time is between 1 day and 1 month
return $dtF->diff($dtT)->format('%d days, %h hours');
break;
case ($seconds<60*60*24*365 && $seconds>60*60*24*31): // if time between 1 month and 1 year
return $dtF->diff($dtT)->format('%m months, %d days');
break;
case ($seconds>60*60*24*365): // if time is longer than 1 year
return $dtF->diff($dtT)->format('%y years, %m months');
break;
}
答案 16 :(得分：0)
@Glavić答案的变体-这个隐藏了前导零以缩短结果，并在正确的位置使用了复数。它还会消除不必要的精度（例如，如果时差超过2小时，那么您可能不在乎是多少分钟或几秒钟）。
function secondsToTime($seconds)
{
$dtF = new \DateTime('@0');
$dtT = new \DateTime("@$seconds");
$dateInterval = $dtF->diff($dtT);
$days_t = 'day';
$hours_t = 'hour';
$minutes_t = 'minute';
$seconds_t = 'second';
if ((int)$dateInterval->d > 1) {
$days_t = 'days';
}
if ((int)$dateInterval->h > 1) {
$hours_t = 'hours';
}
if ((int)$dateInterval->i > 1) {
$minutes_t = 'minutes';
}
if ((int)$dateInterval->s > 1) {
$seconds_t = 'seconds';
}
if ((int)$dateInterval->d > 0) {
if ((int)$dateInterval->d > 1 || (int)$dateInterval->h === 0) {
return $dateInterval->format("%a $days_t");
} else {
return $dateInterval->format("%a $days_t, %h $hours_t");
}
} else if ((int)$dateInterval->h > 0) {
if ((int)$dateInterval->h > 1 || (int)$dateInterval->i === 0) {
return $dateInterval->format("%h $hours_t");
} else {
return $dateInterval->format("%h $hours_t, %i $minutes_t");
}
} else if ((int)$dateInterval->i > 0) {
if ((int)$dateInterval->i > 1 || (int)$dateInterval->s === 0) {
return $dateInterval->format("%i $minutes_t");
} else {
return $dateInterval->format("%i $minutes_t, %s $seconds_t");
}
} else {
return $dateInterval->format("%s $seconds_t");
}
}
php > echo secondsToTime(60);
1 minute
php > echo secondsToTime(61);
1 minute, 1 second
php > echo secondsToTime(120);
2 minutes
php > echo secondsToTime(121);
2 minutes
php > echo secondsToTime(2000);
33 minutes
php > echo secondsToTime(4000);
1 hour, 6 minutes
php > echo secondsToTime(4001);
1 hour, 6 minutes
php > echo secondsToTime(40001);
11 hours
php > echo secondsToTime(400000);
4 days
答案 17 :(得分：0)
我正在编辑其中一个代码，以便在出现负值时正常工作。当值为负数时，floor()函数未给出正确的计数。因此，我们需要先在abs()函数中使用floor()函数。
$inputSeconds变量可以是当前时间戳和所需日期之间的差。
/**
* 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
$days = abs($inputSeconds / $secondsInADay);
$days = floor($days);
// extract hours
$hourSeconds = $inputSeconds % $secondsInADay;
$hours = abs($hourSeconds / $secondsInAnHour);
$hours = floor($hours);
// extract minutes
$minuteSeconds = $hourSeconds % $secondsInAnHour;
$minutes = abs($minuteSeconds / $secondsInAMinute);
$minutes = floor($minutes);
// extract the remaining seconds
$remainingSeconds = $minuteSeconds % $secondsInAMinute;
$seconds = abs($remainingSeconds);
$seconds = ceil($remainingSeconds);
// return the final array
$obj = array(
'd' => (int) $days,
'h' => (int) $hours,
'm' => (int) $minutes,
's' => (int) $seconds,
);
return $obj;
}
答案 18 :(得分：0)
a=int(input("Enter your number by seconds "))
d=a//(24*3600) #Days
h=a//(60*60)%24 #hours
m=a//60%60 #minutes
s=a%60 #seconds
print("Days ",d,"hours ",h,"minutes ",m,"seconds ",s)
答案 19 :(得分：0)
我使用的这个解决方案（可以追溯到学习PHP的时候），没有任何功能上的作用：
sudo apt install libcurl4-openssl-dev尽管这是一个古老的问题，但是遇到这个问题的新学习者可能会发现此答案很有用。
答案 20 :(得分：0)
使用DateInterval：
$d1 = new DateTime();
$d2 = new DateTime();
$d2->add(new DateInterval('PT'.$timespan.'S'));
$interval = $d2->diff($d1);
echo $interval->format('%a days, %h hours, %i minutes and %s seconds');
// Or
echo sprintf('%d days, %d hours, %d minutes and %d seconds',
$interval->days,
$interval->h,
$interval->i,
$interval->s
);
// $interval->y => years
// $interval->m => months
// $interval->d => days
// $interval->h => hours
// $interval->i => minutes
// $interval->s => seconds
// $interval->days => total number of days
答案 21 :(得分：0)
我写的Interval类可以使用。它也可以以相反的方式使用。
composer require lubos/cakephp-interval
$Interval = new \Interval\Interval\Interval();
// output 2w 6h
echo $Interval->toHuman((2 * 5 * 8 + 6) * 3600);
// output 36000
echo $Interval->toSeconds('1d 2h');
答案 22 :(得分：0)
以下是我喜欢使用的一些代码，用于获取两个日期之间的持续时间。它接受两个日期，并为您提供一个很好的句子结构回复。
这是找到here代码的略微修改版本。
<?php
function dateDiff($time1, $time2, $precision = 6, $offset = false) {
// If not numeric then convert texts to unix timestamps
if (!is_int($time1)) {
$time1 = strtotime($time1);
}
if (!is_int($time2)) {
if (!$offset) {
$time2 = strtotime($time2);
}
else {
$time2 = strtotime($time2) - $offset;
}
}
// If time1 is bigger than time2
// Then swap time1 and time2
if ($time1 > $time2) {
$ttime = $time1;
$time1 = $time2;
$time2 = $ttime;
}
// Set up intervals and diffs arrays
$intervals = array(
'year',
'month',
'day',
'hour',
'minute',
'second'
);
$diffs = array();
// Loop thru all intervals
foreach($intervals as $interval) {
// Create temp time from time1 and interval
$ttime = strtotime('+1 ' . $interval, $time1);
// Set initial values
$add = 1;
$looped = 0;
// Loop until temp time is smaller than time2
while ($time2 >= $ttime) {
// Create new temp time from time1 and interval
$add++;
$ttime = strtotime("+" . $add . " " . $interval, $time1);
$looped++;
}
$time1 = strtotime("+" . $looped . " " . $interval, $time1);
$diffs[$interval] = $looped;
}
$count = 0;
$times = array();
// Loop thru all diffs
foreach($diffs as $interval => $value) {
// Break if we have needed precission
if ($count >= $precision) {
break;
}
// Add value and interval
// if value is bigger than 0
if ($value > 0) {
// Add s if value is not 1
if ($value != 1) {
$interval.= "s";
}
// Add value and interval to times array
$times[] = $value . " " . $interval;
$count++;
}
}
if (!empty($times)) {
// Return string with times
return implode(", ", $times);
}
else {
// Return 0 Seconds
}
return '0 Seconds';
}
答案 23 :(得分：0)
一体化解决方案。没有带零的单位。只生成您指定的单位数（默认为3）。 相当长，也许不是很优雅。定义是可选的，但在大型项目中可能派上用场。
define('OneMonth', 2592000);
define('OneWeek', 604800);
define('OneDay', 86400);
define('OneHour', 3600);
define('OneMinute', 60);
function SecondsToTime($seconds, $num_units=3) {
$time_descr = array(
"months" => floor($seconds / OneMonth),
"weeks" => floor(($seconds%OneMonth) / OneWeek),
"days" => floor(($seconds%OneWeek) / OneDay),
"hours" => floor(($seconds%OneDay) / OneHour),
"mins" => floor(($seconds%OneHour) / OneMinute),
"secs" => floor($seconds%OneMinute),
);
$res = "";
$counter = 0;
foreach ($time_descr as $k => $v) {
if ($v) {
$res.=$v." ".$k;
$counter++;
if($counter>=$num_units)
break;
elseif($counter)
$res.=", ";
}
}
return $res;
}
随意投票，但请务必在代码中尝试。它可能就是你所需要的。
答案 24 :(得分：-1)
foreach ($email as $temp => $value) {
$dat = strtotime($value['subscription_expiration']); //$value come from mysql database
//$email is an array from mysqli_query()
$date = strtotime(date('Y-m-d'));
$_SESSION['expiry'] = (((($dat - $date)/60)/60)/24)." Days Left";
//you will get the difference from current date in days.
}
答案 25 :(得分：-2)
这是我过去用来从与您的问题相关的另一个日期中减去日期的功能，我的原则是在产品过期之前剩下多少天，小时分钟和秒：
$expirationDate = strtotime("2015-01-12 20:08:23");
$toDay = strtotime(date('Y-m-d H:i:s'));
$difference = abs($toDay - $expirationDate);
$days = floor($difference / 86400);
$hours = floor(($difference - $days * 86400) / 3600);
$minutes = floor(($difference - $days * 86400 - $hours * 3600) / 60);
$seconds = floor($difference - $days * 86400 - $hours * 3600 - $minutes * 60);
echo "{$days} days {$hours} hours {$minutes} minutes {$seconds} seconds";