mktime

(PHP 4, PHP 5)

mktimeRestituisce la UNIX timestamp per una data

Descrizione

int mktime ( int $hour , int $minute , int $second , int $month , int $day , int $year [, int $is_dst ] )

Attenzione: Nota lo strano ordine degli argomenti, che differiscono dal normale ordine degli argomenti in una normale chiamata UNIX mktime() e che non si presta bene a far comparire i parametri da destra a sinistra (guarda sotto). E' un comune errore la confusione di questi argomenti in uno script.

Restituisce la Unix timestamp corrispondente all'argomento dato. Questa timestamp è un intero lungo contenente il numero di secondi tra la Unix Epoch (January 1 1970) e la data e orario specificati.

Gli argomenti possono essere omessi nell'ordine da destra a sinistra; degli argomenti omessi saranno impostati con l'attuale valore accordandolo alla data e orario locale.

is_dst può essere impostato su 1 se l'orario è nell'ora legale, 0 altrimenti, o -1 (di default) se è sconosciuta la presenza dell'ora legale o meno. Se è sconosciuto, il PHP proverà ad impostarlo da se. Questo può causare un risultato non aspettato (ma non sbagliato).

Nota:

is_dst è stato aggiunto nella verisone 3.0.10.

mktime() è usata per fare calcoli tra date e validazioni, come può calcolare automaticamente il corretto valore per un valore fuori dall'intervallo valido. Per esempio, ognuna delle seguenti linee produce la stringa "Jan-01-1998".

Example #1 Esempio di mktime()

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));
Year può avere sia 2 che 4 cifre, con valori compresi tra 0-69 e 2000-2069 oppure tra 70-99 e 1970-1999 (sui sistemi dove time_t è un intero segnato a 32bit, come sulla maggior parte dei PC di oggi, l'intervallo valido per year è tra 1902 e 2037).

L'ultimo giorno del mese dato può essere espresso come il giorno "0" del mese successivo, non come il giorno -1. Entrami i seguenti esempi produrranno la stringa "L'ultimo giorno di Feb 2000 è: 29".

Example #2 L'ultimo giorno del mese successivo

$lastday = mktime (0,0,0,3,0,2000);
echo strftime ("L'ultimo giorno di Feb 2000 è: %d", $lastday);
     
$lastday = mktime (0,0,0,4,-31,2000);
echo strftime ("L'ultimo giorno di Feb 2000 è: %d", $lastday);

Date con anno, mese e giorno uguali a 0 non sono considerate valide (altrimenti saranno considerate come 30.11.1999, quando hanno uno strano behavior).

Guarda anche date() e time().

add a note add a note

User Contributed Notes 21 notes

up
6
joseph dot andrew dot hughes at gmail dot com
6 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.)
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6
info at microweb dot lt
3 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.]
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4
Alan
5 years ago
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.
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4
tom at chegg dot com
3 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) . ",";
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1
ronnie dot kurniawan at gmail dot com
5 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;
}
?>
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1
thomas_corthals at hotmail dot com
5 years ago
It seems mktime() doesn't return negative timestamps on Linux systems with a version of glibc <= 2.3.3.
up
0
Rad
24 days 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
?>
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0
mogster at redesign dot no
1 month 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));
}
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0
delfino dot salinas at gmail dot com
1 month 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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-1
ionut dot bodea at eydos dot ro
5 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
)
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-1
ooogla at hotmail dot com
5 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.
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-2
rlz
6 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.
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-1
PHPcoder at freemail dot ig3 dot net
6 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().
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-2
Stephen
7 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
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-2
yan
5 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;

?>
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-3
cebleo at n-trance dot net
4 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.
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-2
rga at merchantpal dot com
7 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), 312007 );
$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 - $d_dayofyear);                // how many days left in year
?>
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-2
info at djdb dot be
7 months ago
raw date to clean timestamp
 private function dateToTimestamp($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]);
        }
    }
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-3
Jacob Santos
1 year 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.
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-3
zfowler at unomaha dot edu
4 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.
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-5
contact at phpmember dot com
4 years ago
How many days have  passed since the beginning of the year.... regardless of what year it is...

<?php
//Carlos Galindo
//phpmember.com

$days = floor((time()-mktime(null,null,null,1,0,date("Y")))/86400);
           
echo
"$days days have passed";

//Good Luck
?>
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