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.
mktime
(PHP 4, PHP 5, PHP 7, PHP 8)
mktime — 日付を Unix のタイムスタンプとして取得する
説明
mktime(
int
?int
?int
?int
?int
?int
): int|false
int
$hour,?int
$minute = null,?int
$second = null,?int
$month = null,?int
$day = null,?int
$year = null): int|false
与えられた引数に従って UNIX のタイムスタンプを返します。 このタイムスタンプは、Unix epoch(1970年1月1日00:00:00 GMT)から 指定された時刻までの通算秒を表す長整数です。
省略されたり、null が指定されたオプションの引数は、
ローカルの日付と時刻に従って、現在の値にセットされます。
警告
引数の順番が変なので注意して下さい:
month, day,
year の順になっており、
理にかなった year, month,
day の順になっていません。
引数なしで mktime() を呼び出すことはサポートされていません。 引数なしで呼び出すと、ArgumentCountError がスローされます。 現在のタイムスタンプを取得する目的には、time() が使えます。
パラメータ
-
hour -
month、dayとyearで決まる日付の 0 時から数えた「時」。負の値は、その日の 0 時から前にさかのぼった時間を表します。 23 より大きい値は、その翌日以降の該当する時間を表します。 -
minute -
hour時 0 分から数えた「分」。 負の値は、その前の時刻を表します。 59 より大きい値は、その次の時間以降の該当する時間を表します。 -
second -
minute分 0 秒から数えた「秒」。 負の値は、その前の時刻を表します。 59 より大きい値は、その次の分以降の該当する時間を表します。 -
month -
前年末から数えた月数。1 から 12 までの場合は、カレンダーどおりのその年の「月」を表します。 (負の値を含めた) 1 より小さい値は、前年の月を逆順でたどります。 つまり 0 なら 12 月、-1 なら 11 月になるということです。 12 より大きい値は、その翌年以降の該当する月を表します。
-
day -
前月末から数えた日数。1 から 28、29、30、31 (月によって異なる) までの場合は、その月の「日」を表します。 (負の値を含めた) 1 より小さい値は、前月の日を逆順でたどります。 つまり 0 なら前月の末日、-1 ならそのさらに前日になるということです。 その月の日数より大きい値は、翌月以降の該当する日を表します。
-
year -
年。2 桁または 4 桁の値を指定可能で、 0-69 の間の値は 2000-2069 に、70-100 は 1970-2000 にマップされます。 今日最も一般的なシステム、すなわち time_t が 32 ビットの符号付き整数である システムでは
yearとして有効な範囲は 1901 から 2038 の間です。
戻り値
mktime() は与えられた引数の Unix
タイムスタンプを返します。
タイムスタンプの値が PHP の整数型に合わない場合は、false を返します。
変更履歴
| バージョン | 説明 |
|---|---|
| 8.0.0 |
hour は、オプションではなくなりました。
Unixタイムスタンプ が必要な場合、time()
を使います。
|
| 8.0.0 |
minute, second, month,
day, year は、nullable になりました。
|
例
例1 mktime() の基本的な例
<?php
// デフォルトのタイムゾーンを設定します。
date_default_timezone_set('UTC');
// 出力: July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", mktime(0, 0, 0, 7, 1, 2000));
// 出力例: 2006-04-05T01:02:03+00:00
echo date('c', mktime(1, 2, 3, 4, 5, 2006));
?>
例2 mktime() の例
mktime() は入力日付の有効性を確認しており、 範囲外の入力を自動的に修正して計算してくれるので便利です。 例えば、以下の各行はいずれも文字列 "Jan-01-1998"を出力します。
<?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));
?>
例3 月の末日
指定した月の最後の日は次の月の -1 番目の日ではなく、"0" 番目の日 として表すことが可能です。以下の例はともに文字列"The last day in Feb 2000 is: 29" を出力します。
<?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);
?>参考
- DateTimeImmutable
- checkdate() - グレゴリオ暦の日付/時刻の妥当性を確認します
- gmmktime() - GMT 日付から Unix タイムスタンプを取得する
- date() - Unixタイムスタンプを書式化する
- time() - 現在の Unix タイムスタンプを返す
add a note
User Contributed Notes 27 notes
Alan ¶
15 years ago
Rad ¶
10 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 ¶
16 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 ¶
6 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 ¶
7 years ago
Please mind function is timezone dependent. Timezone independent funciton is gmmktime
thomas_corthals at hotmail dot com ¶
15 years ago
It seems mktime() doesn't return negative timestamps on Linux systems with a version of glibc <= 2.3.3.
MarkAgius at markagius dot co dot uk ¶
6 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);
}
?>
mh240873 at web dot de ¶
9 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 $day reaches 2015-10-25 (end of summer time in Germany):
24.10.2015
25.10.2015
25.10.2015 // Ups! Same date twice in calendar
27.10.2015
You may workaround this by using date_default_timezone_set('UTC') where all days have the same number of seconds.
tom at chegg dot com ¶
13 years ago
I was using the following to get a list of month names.
for ($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,$i,1) . ",";
info at microweb dot lt ¶
13 years ago
Function to generate array of dates between two dates (date range array)
<?php
function dates_range($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 $dates_range;
}
echo '<pre>';
print_r(dates_range('2009-12-25', '2010-01-05'));
echo '</pre>';
?>
[EDIT BY danbrown AT php DOT net: Contains a bugfix submitted by (carlosbuz2 AT gmail DOT com) on 04-MAR-2011, with the following note: The first date in array is incorrect.]
ronnie dot kurniawan at gmail dot com ¶
15 years ago
Add (and subtract) unixtime:
<?php
function utime_add($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 ¶
16 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 ¶
17 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 ¶
7 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
?>
Stephen ¶
17 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
info at djdb dot be ¶
10 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]);
}
}
yan ¶
15 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 ¶
14 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.
cebleo at n-trance dot net ¶
14 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', $hours_diff);
?>
it shows: 01:00 Hours.
rlz ¶
16 years ago
Finding out the number of days in a given month and year, accounting for leap years when February has more than 28 days.
<?php
function days_in_month($year, $month) {
return( date( "t", mktime( 0, 0, 0, $month, 1, $year) ) );
}
?>
Hope it helps a soul out there.
ooogla at hotmail dot com ¶
15 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']))));
$count is incremented by the loop.
mogster at redesign dot no ¶
10 years ago
Just a simple function to return mktime from a db (mysql) datetime (Y-m-d H:i:s):
function retMktimest($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 ¶
7 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 || $q == 4 ? 31 : 30);
}
Jacob Santos ¶
11 years ago
Please note that incrementing a date using mktime in a loop is not proper. You could do it, except that there is a far better method found in the DateTime PHP class. Look at the documentation for DateTime::modify, DateTime::add (when supported) and DateTime::sub (when supported).
Also, adding seconds to a time is, well it isn't as easy as it seems, "Hey I'll just add 3600 seconds or 86400 seconds or x seconds!". The phrase once bitten, twice shy is quite applicable with the usage of adding seconds. If you ever had to 'fix' a time by calculating midnight to add the correct number of seconds, then you are doing it wrong.
Luckily, knowing is not a requirement, because DateTime and friends exists, removing the complexity for you.
So if given a choice of
mktime($seconds, $minutes, $hours+1);
and
$datetime->modify('+1 hour');
or
$datetime->add('P1H');
I'll go with the second choice, but probably not the third, unless I was using DateInterval::createFromDateString, so that other developers knew my intent.
mktime_php at mailinator dot com ¶
8 years ago
One practical and useful example of using negative values in mktime is the following:
<?php
//Considering today's date
echo date('Y-m-d'); //Prints: 2016-03-22
echo date('Y-m-d', mktime(0, 0, 0, date("m"), date("d")-42, date("Y"))); //Prints: 2016-02-09
?>
By using date outputs inside mktime and adding or subtracting from them may be simpler than using other methods (string concatenations or timestamp values) and less prone to human calculations' errors.
ionut dot bodea at eydos dot ro ¶
15 years ago
Here is what I use to calculate age. It took me 30 minutes to write and it's quite accurate. What it has special is that it's calculating the number of days a year has (float number), by testing if a year is a leap one or not. This number is used to compute the age.
<?php
function get_age($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 ¶
10 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);
}
