Do remember that, counter-intuitively enough, the arguments for month and day are inversed (or middle-endian). A common mistake for Europeans seems to be to feed the date arguments in the expected order (big endian or little endian).
It's clear to see where this weird order comes from (even with the date being big endian the order for all arguments would still be mixed - it's obviously based on the American date format with the time "prefixed" to allow an easier shorthand) and why this wasn't changed (passing the values in the wrong order produces a valid, though unexpected, result in most cases), but it continues to be a source of confusion for me whenever I come back to PHP from other languages or libraries.
PHP - Manual: mktime
2024-12-22
mktime
(PHP 4, PHP 5, PHP 7, PHP 8)
mktime — 取得一个日期的 Unix 时间戳
说明
mktime(
int
int
int
int
int
int
int
): int
int
$hour = date("H"),int
$minute = date("i"),int
$second = date("s"),int
$month = date("n"),int
$day = date("j"),int
$year = date("Y"),int
$is_dst = -1): int
根据给出的参数返回 Unix 时间戳。时间戳是一个长整数，包含了从 Unix 纪元（January 1 1970 00:00:00 GMT）到给定时间的秒数。
参数可以从右向左省略，任何省略的参数会被设置成本地日期和时间的当前值。
注释
注意:
As of PHP 5.1, when called with no arguments, mktime() throws an
E_STRICTnotice: use the time() function instead.
参数
-
hour -
小时数。 The number of the hour relative to the start of the day determined by
month,dayandyear. Negative values reference the hour before midnight of the day in question. Values greater than 23 reference the appropriate hour in the following day(s). -
minute -
分钟数。 The number of the minute relative to the start of the
hour. Negative values reference the minute in the previous hour. Values greater than 59 reference the appropriate minute in the following hour(s). -
second -
秒数（一分钟之内）。 The number of seconds relative to the start of the
minute. Negative values reference the second in the previous minute. Values greater than 59 reference the appropriate second in the following minute(s). -
month -
月份数。 The number of the month relative to the end of the previous year. Values 1 to 12 reference the normal calendar months of the year in question. Values less than 1 (including negative values) reference the months in the previous year in reverse order, so 0 is December, -1 is November, etc. Values greater than 12 reference the appropriate month in the following year(s).
-
day -
天数。 The number of the day relative to the end of the previous month. Values 1 to 28, 29, 30 or 31 (depending upon the month) reference the normal days in the relevant month. Values less than 1 (including negative values) reference the days in the previous month, so 0 is the last day of the previous month, -1 is the day before that, etc. Values greater than the number of days in the relevant month reference the appropriate day in the following month(s).
-
year -
年份数，可以是两位或四位数字，0-69 对应于 2000-2069，70-100 对应于 1970-2000。在如今系统中普遍把 time_t 作为一个 32 位有符号整数的情况下，
year的合法范围是 1901 到 2038 之间，不过此限制自 PHP 5.1.0 起已被克服了。 -
is_dst -
本参数可以设为 1，表示正处于夏时制时间（DST），0 表示不是夏时制，或者 -1（默认值）表示不知道是否是夏时制。如果未知，PHP 会尝试自己搞明白。这可能产生不可预知（但并非不正确）的结果。如果 PHP 运行的系统中启用了 DST 或者
is_dst设为 1，某些时间是无效的。例如 DST 自 2:00 生效，则所有处于 2:00 到 3:00 之间的时间都无效，mktime() 会返回一个未定义（通常为负）的值。某些系统（例如 Solaris 8）的 DST 在午夜生效，则 DST 生效当天的 0:30 会被计算为前一天的 23:30。注意:
自 PHP 5.1.0 起，本参数已被废弃。应该使用新的时区处理特性来替代。
注意:
PHP 7.0.0 起，此参数已经被移除。
返回值
mktime() 根据给出的参数返回 Unix
时间戳。如果参数非法，本函数返回
false（在 PHP 5.1 之前返回 -1）。
错误／异常
在每次调用日期/时间函数时，如果时区无效则会引发 E_NOTICE 错误。参见
date_default_timezone_set()。
更新日志
| 版本 | 说明 |
|---|---|
| 7.0.0 |
is_dst参数已经被移除。
|
| 5.3.0 |
mktime() now throws E_DEPRECATED notice
if the is_dst parameter is used.
|
| 5.1.0 |
is_dst 参数被废弃。出错时函数返回
false 而不再是 -1。修正了本函数可以接受年月日参数全为零。
|
| 5.1.0 |
When called with no arguments, mktime() throws
E_STRICT notice. Use the
time() function instead.
|
| 5.1.0 |
现在发布 |
范例
示例 #1 基本例子
<?php
// Set the default timezone to use. Available as of PHP 5.1
date_default_timezone_set('UTC');
// Prints: July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", mktime(0, 0, 0, 7, 1, 2000));
// Prints something like: 2006-04-05T01:02:03+00:00
echo date('c', mktime(1, 2, 3, 4, 5, 2006));
?>
示例 #2 mktime() 例子
mktime() 在做日期计算和验证方面很有用，它会自动计算超出范围的输入的正确值。例如下面例子中每一行都会产生字符串 "Jan-01-1998"。
<?php
echo date("M-d-Y", mktime(0, 0, 0, 12, 32, 1997));
echo date("M-d-Y", mktime(0, 0, 0, 13, 1, 1997));
echo date("M-d-Y", mktime(0, 0, 0, 1, 1, 1998));
echo date("M-d-Y", mktime(0, 0, 0, 1, 1, 98));
?>
示例 #3 下个月的最后一天
任何给定月份的最后一天都可以被表示为下个月的第 "0" 天，而不是 -1 天。下面两个例子都会产生字符串 "The last day in Feb 2000 is: 29"。
<?php
$lastday = mktime(0, 0, 0, 3, 0, 2000);
echo strftime("Last day in Feb 2000 is: %d", $lastday);
$lastday = mktime(0, 0, 0, 4, -31, 2000);
echo strftime("Last day in Feb 2000 is: %d", $lastday);
?>
注释
警告
在 PHP 5.1.0 之前，在任何已知 Windows 版本以及一些其它系统下不支持负的时间戳。因此年份的有效范围限制为 1970 到 2038。
参见
- checkdate() - 验证一个格里高里日期
- gmmktime() - 取得 GMT 日期的 UNIX 时间戳
- date() - 格式化一个本地时间／日期
- time() - 返回当前的 Unix 时间戳
add a note
User Contributed Notes 27 notes
Alan ¶
13 years ago
Rad ¶
8 years ago
Be careful passing zeros into mktime, in most cases a zero will count as the previous unit of time. The documentation explains this yet most of the comments here still use zeroes.
For example, if you pass the year 2013 into mktime, with zeroes for everything else, the outcome is probably not what you are looking for.
<?php
echo date('F jS, Y g:i:s a', mktime(0, 0, 0, 0, 0, 2013));
// November 30th, 2012 12:00:00 am
?>
Instead of using 0's, try 1's. This makes more sense (except for minutes/seconds). Maybe not as obvious of a purpose as zeroes to other programmers, though.
<?php
echo date('F jS, Y g:i:s a', mktime(1, 1, 1, 1, 1, 2013));
// January 1st, 2013 1:01:01 am
?>
joseph dot andrew dot hughes at gmail dot com ¶
14 years ago
Just a small thing to think about if you are only trying to pull the month out using mktime and date. Make sure you place a 1 into day field. Otherwise you will get incorrect dates when a month is followed by a month with less days when the day of the current month is higher then the max day of the month you are trying to find.. (Such as today being Jan 30th and trying to find the month Feb.)
developers at zeros dot co dot id ¶
4 years ago
Please note, mktime requires an integer value, if you use date("H"), date("i"), date("s") as a value, which is actually have a leading zero, you may get "A non well formed numeric value encountered" notice. so you need some tricks like this
mktime( date("G"), intval(date("i")), intval(date("s"), date("n"), date("j"), date("Y") )
Since there are no minute & second without leading zero in the date function, we can use the intval() function or you can cast value type like this to force the value type.
(int) date("i")
phper ¶
6 years ago
Please mind function is timezone dependent. Timezone independent funciton is gmmktime
MarkAgius at markagius dot co dot uk ¶
5 years ago
The following function moves all the parameters in order of most significant (biggest) to least significant (smallest) order.
Year is bigger than month. Month is bigger than day. Day bigger than hours...
Much less confusing than mktime order.
<?php
function mkTimestamp($year,$month,$day, $hours=0,$minutes=0,$seconds=0){
// Same as mktime() but parameters are in most significant to least significant order.
return mktime($hours,$minutes,$seconds, $month,$day,$year);
}
?>
thomas_corthals at hotmail dot com ¶
14 years ago
It seems mktime() doesn't return negative timestamps on Linux systems with a version of glibc <= 2.3.3.
mh240873 at web dot de ¶
7 years ago
Pay attention that not all days have the same number of seconds (86400s) if you are using date_default_timezone_set(..) and the used timezone has Daylight Saving Time (DST) e.g. "Europe/Berlin". Under PHP 5.5.16 I get the following results:
$shortday = mktime(23,59,59, '3','29','2015') - mktime(0,0,0, '3','29','2015) + 1; // result: 82800s (86400s - 3600s)
$normalDay = mktime(23,59,59, '1', '2','2015') - mktime(0,0,0, '1', '1','2015) + 1; // result: 86400s
$longDay = mktime(23,59,59,'10','25','2015') - mktime(0,0,0,'10','25','2015) + 1; // result: 90000s (86400s + 3600s)
Pitfall is noticeable if you are running an iterative loop with a code like:
echo date( 'd.m.Y', $day );
$day = $day + 86400; // 86400 = 24*3600 - frequently used in PHP code
which results in wrong date if $day reaches 2015-10-25 (end of summer time in Germany):
24.10.2015
25.10.2015
25.10.2015 // Ups! Same date twice in calendar
27.10.2015
You may workaround this by using date_default_timezone_set('UTC') where all days have the same number of seconds.
tom at chegg dot com ¶
11 years ago
I was using the following to get a list of month names.
for ($i=1; $i<13; $i++) {
echo date('F', mktime(0,0,0,$i) . ",";
}
Normally this outputs -
January,February,March,April,May,June,July,August,
September,October,November,December
However if today's date is the 31st you get instead:
January,March,March,May,May,July,July,August,October,
October,December,December
Why? Because Feb,Apr,June,Sept, and Nov don't have 31 days!
The fix, add the 5th parameter, don't let the day of month default to today's date:
echo date('F', mktime(0,0,0,$i,1) . ",";
info at microweb dot lt ¶
11 years ago
Function to generate array of dates between two dates (date range array)
<?php
function dates_range($date1, $date2)
{
if ($date1<$date2)
{
$dates_range[]=$date1;
$date1=strtotime($date1);
$date2=strtotime($date2);
while ($date1!=$date2)
{
$date1=mktime(0, 0, 0, date("m", $date1), date("d", $date1)+1, date("Y", $date1));
$dates_range[]=date('Y-m-d', $date1);
}
}
return $dates_range;
}
echo '<pre>';
print_r(dates_range('2009-12-25', '2010-01-05'));
echo '</pre>';
?>
[EDIT BY danbrown AT php DOT net: Contains a bugfix submitted by (carlosbuz2 AT gmail DOT com) on 04-MAR-2011, with the following note: The first date in array is incorrect.]
ronnie dot kurniawan at gmail dot com ¶
13 years ago
Add (and subtract) unixtime:
<?php
function utime_add($unixtime, $hr=0, $min=0, $sec=0, $mon=0, $day=0, $yr=0) {
$dt = localtime($unixtime, true);
$unixnewtime = mktime(
$dt['tm_hour']+$hr, $dt['tm_min']+$min, $dt['tm_sec']+$sec,
$dt['tm_mon']+1+$mon, $dt['tm_mday']+$day, $dt['tm_year']+1900+$yr);
return $unixnewtime;
}
?>
PHPcoder at freemail dot ig3 dot net ¶
14 years ago
The maximum possible date accepted by mktime() and gmmktime() is dependent on the current location time zone.
For example, the 32-bit timestamp overflow occurs at 2038-01-19T03:14:08+0000Z. But if you're in a UTC -0500 time zone (such as EST in North America), the maximum accepted time before overflow (for older PHP versions on Windows) is 2038-01-18T22:14:07-0500Z, regardless of whether you're passing it to mktime() or gmmktime().
rga at merchantpal dot com ¶
15 years ago
You cannot simply subtract or add month VARs using mktime to obtain previous or next months as suggested in previous user comments (at least not with a DD > 28 anyway).
If the date is 03-31-2007, the following yeilds March as a previous month. Not what you wanted.
<?php
$dateMinusOneMonth = mktime(0, 0, 0, (3-1), 31, 2007 );
$lastmonth = date("n | F", $dateMinusOneMonth);
echo $lastmonth; //---> 3 | March
?>
mktime correctly gives you back the 3rd of March if you subtract 1 month from March 31 (there are only 28 days in Feb 07).
If you are just looking to do month and year arithmetic using mktime, you can use general days like 1 or 28 to do stuff like this:
<?php
$d_daysinmonth = date('t', mktime(0,0,0,$myMonth,1,$myYear)); // how many days in month
$d_year = date('Y', mktime(0,0,0,$myMonth,1,$myYear)); // year
$d_isleapyear = date('L', mktime(0,0,0,$myMonth,1,$myYear)); // is YYYY a leapyear?
$d_firstdow = date('w', mktime(0,0,0,$myMonth,'1',$myYear)); // FIRST falls on what day of week (0-6)
$d_firstname = date('l', mktime(0,0,0,$myMonth,'1',$myYear)); // FIRST falls on what day of week Full Name
$d_month = date('n', mktime(0,0,0,$myMonth,28,$myYear)); // month of year (1-12)
$d_monthname = date('F', mktime(0,0,0,$myMonth,28,$myYear)); // Month Long name (July)
$d_month_previous = date('n', mktime(0,0,0,($myMonth-1),28,$myYear)); // PREVIOUS month of year (1-12)
$d_monthname_previous = date('F', mktime(0,0,0,($myMonth-1),28,$myYear)); // PREVIOUS Month Long name (July)
$d_month_next = date('n', mktime(0,0,0,($myMonth+1),28,$myYear)); // NEXT month of year (1-12)
$d_monthname_next = date('F', mktime(0,0,0,($myMonth+1),28,$myYear)); // NEXT Month Long name (July)
$d_year_previous = date('Y', mktime(0,0,0,$myMonth,28,($myYear-1))); // PREVIOUS year
$d_year_next = date('Y', mktime(0,0,0,$myMonth,28,($myYear+1))); // NEXT year
$d_weeksleft = (52 - $d_weekofyear); // how many weeks left in year
$d_daysinyear = $d_isleapyear ? 366 : 365; // set correct days in year for leap years
$d_daysleft = ($d_daysinyear - $d_dayofyear); // how many days left in year
?>
A.Ross ¶
5 years ago
What's odd is that mktime doesn't seem to support every possible year number. It's common sense that 2 digit (shortened) year numbers are interpreted in the range 1970..2069
However, when padded with zeroes, no such transformation should happen (at least that is the behaviour of other date functions). Unfortunately it does (until year 100 *inclusive*):
<?php
echo date("Y-m-d",mktime(0,0,0,1,1,"0001"));
// Expected: 0001-01-01
// Result: 2001-01-01 INCORRECT
echo date("Y-m-d",mktime(0,0,0,1,1,"0100"));
// Expected: 0100-01-01
// Result: 2000-01-01 INCORRECT
echo date("Y-m-d",mktime(0,0,0,1,1,"0101"));
// Expected: 0101-01-01
// Result: 0101-01-01 Correct
?>
info at djdb dot be ¶
8 years ago
raw date to clean timestamp
private function dateToTimestamp($date){
$datefrom = explode(" ", $date);
$value = array();
if(strpos($datefrom[0], '-')){
//print "issplit -";
$value = explode("-", $datefrom[0]);
}
if(strpos($datefrom[0], '/')){
//print "issplit /";
$value = explode("/", $datefrom[0]);
}
/*if(){
}*/
if(strlen($value[2])==4){//13/12/2012
//int mktime([hour[minute[second[month[day[year
return mktime(0, 0, 0,$value[1],$value[0],$value[2]);
}else{ //2012/12/13
//int mktime([hour[minute[second[month[day[year
return mktime(0, 0, 0,$value[1],$value[2],$value[0]);
}
}
Stephen ¶
15 years ago
There are several warnings here about using mktime() to determine a date difference because of daylight savings time. However, nobody seems to have mentioned the other obvious problem, which is leap years.
Leap years mean that any effort to use mktime() and time() to determine the age (positive or negative) of some timestamp in years will be flawed. There are some years that are 366 days long, therefore you cannot say that there is a set number of seconds per year.
Timestamps are good for determining *real* time, which is not the same thing as *human calendar* time. The Gregorian calendar is only an approximation of real time, which is tweaked with daylight savings time and leap years to make it conform more to humans' expectations of how time should or ought to work. Timestamps are not tweaked and therefore are the only authoritative way of recording in computers a proper order of succession of events, but they cannot be integrated with a Gregorian system unless you take both leap years and DST into account. Otherwise, you may get the wrong number of years when you are approaching a value of exactly X years.
As for PHP, you could still use timestamps as a way of determining age if you took into account not only DST but also whether or not each year is a leap year and adjusted your calculations accordingly. However, this could become messy and inefficient.
There is an alternative approach to calculating days given the day, month and year of the dates to be compared. Compare the years first, and then compare the month and day - if the month and day have already passed (or, if you like, if they match the current month and day), then add 1 to the total for the years.
This solution works because it stays within the Gregorian system and doesn't venture into the world of timestamps.
There is also the issue of leap seconds, but this will only arise if you literally need to get the *exact* age in seconds. In that case, of course, you would also need to verify that your timestamps are exactly correct and are not delayed by script processing time, plus you would need to determine whether your system conforms to UTC, etc. I expect this will hardly be an issue for anybody using PHP, however if you are interested there is an article on this issue on Wikipedia:
http://en.wikipedia.org/wiki/Leap_second
yan ¶
13 years ago
caculate days between two date
<?php
// end date is 2008 Oct. 11 00:00:00
$_endDate = mktime(0,0,0,11,10,2008);
// begin date is 2007 May 31 13:26:26
$_beginDate = mktime(13,26,26,05,31,2007);
$timestamp_diff= $_endDate-$_beginDate +1 ;
// how many days between those two date
$days_diff = $timestamp_diff/86400;
?>
zfowler at unomaha dot edu ¶
12 years ago
Proper way to convert Excel dates into PHP-friendly timestamps using mktime():
<?php
// The date 6/30/2009 is stored as 39994 in Excel
$days = 39994;
// But you must subtract 1 to get the correct timestamp
$ts = mktime(0,0,0,1,$days-1,1900);
// So, this would then match Excel's representation:
echo date("m/d/Y",$ts);
?>
Excel uses "number of days since Jan. 1, 1900" to store its dates. It also treats 1900 as a leap year when it wasn't, thus there is an extra day which must be accounted for in PHP (and the rest of the world). Subtracting 1 from Excel's number will fix this problem.
rlz ¶
14 years ago
Finding out the number of days in a given month and year, accounting for leap years when February has more than 28 days.
<?php
function days_in_month($year, $month) {
return( date( "t", mktime( 0, 0, 0, $month, 1, $year) ) );
}
?>
Hope it helps a soul out there.
cebleo at n-trance dot net ¶
12 years ago
to ADD or SUBSTRACT times NOTE that if you dont specify the UTC zone your result is the difference +- your server UTC delay.
if you are ina utc/GMT +1
<?php
$hours_diff = strtotime("20:00:00")-strtotime("19:00:00");
echo date('h:i', $hours_diff)." Hours";
?>
it shows: 02:00 Hours
but if you use a default UTC time:
<?php
date_default_timezone_set('UTC');
$hours_diff = strtotime("20:00:00")-strtotime("19:00:00");
echo "<br>". date('h:i', $hours_diff);
?>
it shows: 01:00 Hours.
mogster at redesign dot no ¶
8 years ago
Just a simple function to return mktime from a db (mysql) datetime (Y-m-d H:i:s):
function retMktimest($dbdate) {
return mktime(substr($dbdate, 11, 2), substr($dbdate, 14, 2), substr($dbdate, 17, 2), substr($dbdate, 5, 2), substr($dbdate, 8, 2), substr($dbdate, 0, 4));
}
utilmind ¶
5 years ago
// here is the function which returns the Unix timestamp of last date of quarter, by quarter number:
function last_day_of_quarter($q) {
return mktime(0, 0, 0, floor($q*3), $q == 1 || $q == 4 ? 31 : 30);
}
ooogla at hotmail dot com ¶
13 years ago
If you want to increment the day based on a variable when using a loop you can use this when you submit a form
1. Establish a start date and end date in two different variables
2. Get the number of days between a date
$ndays = (strtotime($_POST['edate']) - strtotime($_POST['sdate'])) / (60 * 60 * 24);
Then here is the string you slip in your loop
$nextday = date('Y-m-d', mktime(0, 0, 0, date("m", strtotime($_POST['sdate'])) , date("d", strtotime($_POST['sdate']))+ $count, date("Y", strtotime($_POST['sdate']))));
$count is incremented by the loop.
Jacob Santos ¶
9 years ago
Please note that incrementing a date using mktime in a loop is not proper. You could do it, except that there is a far better method found in the DateTime PHP class. Look at the documentation for DateTime::modify, DateTime::add (when supported) and DateTime::sub (when supported).
Also, adding seconds to a time is, well it isn't as easy as it seems, "Hey I'll just add 3600 seconds or 86400 seconds or x seconds!". The phrase once bitten, twice shy is quite applicable with the usage of adding seconds. If you ever had to 'fix' a time by calculating midnight to add the correct number of seconds, then you are doing it wrong.
Luckily, knowing is not a requirement, because DateTime and friends exists, removing the complexity for you.
So if given a choice of
mktime($seconds, $minutes, $hours+1);
and
$datetime->modify('+1 hour');
or
$datetime->add('P1H');
I'll go with the second choice, but probably not the third, unless I was using DateInterval::createFromDateString, so that other developers knew my intent.
mktime_php at mailinator dot com ¶
6 years ago
One practical and useful example of using negative values in mktime is the following:
<?php
//Considering today's date
echo date('Y-m-d'); //Prints: 2016-03-22
echo date('Y-m-d', mktime(0, 0, 0, date("m"), date("d")-42, date("Y"))); //Prints: 2016-02-09
?>
By using date outputs inside mktime and adding or subtracting from them may be simpler than using other methods (string concatenations or timestamp values) and less prone to human calculations' errors.
ionut dot bodea at eydos dot ro ¶
13 years ago
Here is what I use to calculate age. It took me 30 minutes to write and it's quite accurate. What it has special is that it's calculating the number of days a year has (float number), by testing if a year is a leap one or not. This number is used to compute the age.
<?php
function get_age($date_start, $date_end) {
$t_lived = get_timestamp($date_end) - get_timestamp($date_start);
$seconds_one_year = get_days_per_year($date_start, $date_end) * 24 * 60 * 60;
$age = array();
$age['years_exact'] = $t_lived / $seconds_one_year;
$age['years'] = floor($t_lived / $seconds_one_year);
$seconds_remaining = $t_lived % $seconds_one_year;
$age['days'] = round($seconds_remaining / (24 * 60 * 60));
return $age;
}
function get_timestamp($date) {
list($y, $m, $d) = explode('-', $date);
return mktime(0, 0, 0, $m, $d, $y);
}
function get_days_per_year($date_start, $date_end) {
list($y1) = explode('-', $date_start);
list($y2) = explode('-', $date_end);
$years_days = array();
for($y = $y1; $y <= $y2; $y++) {
$years_days[] = date('L', mktime(0, 0, 0, 1, 1, $y)) ? 366 : 365;
}
return round(array_sum($years_days) / count($years_days), 2);
}
$date_birth = '1979-10-12';
$date_now = date('Y-m-d');
$age = get_age($date_birth, $date_now);
echo '<pre>';
print_r($age);
echo '</pre>';
?>
It will display something like this:
Array
(
[years_exact] => 28.972974329491
[years] => 28
[days] => 355
)
delfino dot salinas at gmail dot com ¶
8 years ago
this function returns the number of days of a provided month and year, it consider the actual rules for leap years
(if the year is multiple of 4 which is not a multiple of 100 unless multiple of thousand then is a leap)
Regards, hope this function solves any issue :)
function daysinmonth($month,$year) {
$dim = 0;
switch ($month) {
case 1:
case 3:
case 5:
case 7:
case 8:
case 10:
case 12:
$dim=31;
break;
case 4:
case 6:
case 9:
case 11:
$dim=30;
break;
case 2:
if($year%4==0) {
if($year%100==0) {
if($year%1000==0) { $dim=29; } else { $dim=28; }
} else {
$dim=29;
}
} else {$dim=28;}
break;
}
return($dim);
}
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