date

(PHP 4, PHP 5, PHP 7, PHP 8)

date — Форматирует временную метку Unix

Описание

date(string $format, ?int $timestamp = null): string

Возвращает строку, отформатированную в соответствии с указанным в параметре format шаблоном. Используется метка времени, заданная параметром timestamp (метка времени Unix), или текущее системное время, если параметр timestamp не задан. Таким образом, параметр timestamp является необязательным и по умолчанию равен значению, возвращаемому функцией time().

Внимание

Метки времени Unix не обрабатывают часовые пояса. Используйте класс DateTimeImmutable и его метод форматирования DateTimeInterface::format() для форматирования информации о дате/времени с привязкой к часовому поясу.

Список параметров

format: Принятый формат DateTimeInterface::format().

Замечание: Функция date() всегда будет генерировать 000000 в качестве микросекунд, поскольку она принимает параметр int, тогда как DateTime::format() поддерживает микросекунды, если DateTime был создан с микросекундами.
timestamp: Необязательный параметр timestamp — это целочисленная (int) метка времени, по умолчанию равная текущему местному времени, если параметр timestamp не указан или равен null. Говоря по другому, значение по умолчанию равно результату функции time().

Возвращаемые значения

Возвращает отформатированную строку с датой.

Ошибки

Каждый вызов к функциям даты/времени при неправильных настройках часового пояса сгенерирует ошибку уровня E_WARNING, если часовой пояс некорректный. Смотрите также date_default_timezone_set()

Список изменений

Версия	Описание
8.0.0	`timestamp` теперь допускает значение null.

Примеры

Пример #1 Примеры использования функции date()

<?php
// установка часового пояса по умолчанию.
date_default_timezone_set('UTC');


// выведет примерно следующее: Monday
echo date("l");

// выведет примерно следующее: Monday 8th of August 2005 03:12:46 PM
echo date('l jS \of F Y h:i:s A');

// выведет: July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", mktime(0, 0, 0, 7, 1, 2000));

/* пример использования константы в качестве форматирующего параметра */
// выведет примерно следующее: Mon, 15 Aug 2005 15:12:46 UTC
echo date(DATE_RFC822);

// выведет примерно следующее: 2000-07-01T00:00:00+00:00
echo date(DATE_ATOM, mktime(0, 0, 0, 7, 1, 2000));
?>

Чтобы запретить распознавание символа как форматирующего, следует экранировать его с помощью обратного слеша. Если экранированный символ также является форматирующей последовательностью, то следует экранировать его повторно.

Пример #2 Экранирование символов в функции date()

<?php
// выведет примерно следующее: Wednesday the 15th
echo date('l \t\h\e jS');
?>

Для вывода прошедших и будущих дат удобно использовать функции date() и mktime().

Пример #3 Пример совместного использования функций date() и mktime()

<?php
$tomorrow  = mktime(0, 0, 0, date("m")  , date("d")+1, date("Y"));
$lastmonth = mktime(0, 0, 0, date("m")-1, date("d"),   date("Y"));
$nextyear  = mktime(0, 0, 0, date("m"),   date("d"),   date("Y")+1);
?>

Замечание:
Данный способ более надёжен, чем простое вычитание и прибавление секунд к метке времени, поскольку позволяет при необходимости гибко осуществить переход на летнее/зимнее время.

Ещё несколько примеров использования функции date(). Важно отметить, что следует экранировать все символы, которые необходимо оставить без изменений. Это справедливо и для тех символов, которые в текущей версии PHP не распознаются как форматирующие, поскольку это может быть введено в следующих версиях. Для экранировании управляющих последовательностей (например, \n) следует использовать одинарные кавычки.

Пример #4 Форматирование с использованием date()

<?php
// Предположим, что текущей датой является 10 марта 2001, 5:16:18 вечера,
// и мы находимся в часовом поясе Mountain Standard Time (MST)

$today = date("F j, Y, g:i a");                 // March 10, 2001, 5:16 pm
$today = date("m.d.y");                         // 03.10.01
$today = date("j, n, Y");                       // 10, 3, 2001
$today = date("Ymd");                           // 20010310
$today = date('h-i-s, j-m-y, it is w Day');     // 05-16-18, 10-03-01, 1631 1618 6 Satpm01
$today = date('\i\t \i\s \t\h\e jS \d\a\y.');   // it is the 10th day.
$today = date("D M j G:i:s T Y");               // Sat Mar 10 17:16:18 MST 2001
$today = date('H:m:s \m \i\s\ \m\o\n\t\h');     // 17:03:18 m is month
$today = date("H:i:s");                         // 17:16:18
$today = date("Y-m-d H:i:s");                   // 2001-03-10 17:16:18 (формат MySQL DATETIME)
?>

Для форматирования дат на других языках вместо функции date() можно использовать метод IntlDateFormatter::format().

Примечания

Замечание:
Для получения метки времени из строкового представления даты можно воспользоваться функцией strtotime(). Кроме того, некоторые базы данных имеют собственные функции для преобразования внутреннего представления даты в метку времени (например, функция MySQL » UNIX_TIMESTAMP).

Подсказка

Временную метку начала запроса можно получить из поля $_SERVER['REQUEST_TIME'].

Смотрите также

DateTimeImmutable::__construct() - Возвращает новый объект DateTimeImmutable
DateTimeInterface::format() - Возвращает дату, отформатированную согласно переданному формату
gmdate() - Форматирует дату/время по Гринвичу
idate() - Преобразовывает локальное время/дату в целое число
getdate() - Возвращает информацию о дате/времени
getlastmod() - Получает время последней модификации страницы
mktime() - Возвращает метку времени Unix для заданной даты
IntlDateFormatter::format() - Форматирует значение даты/времени в виде строки
time() - Возвращает текущую метку системного времени Unix
Предопределённые константы даты и времени

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User Contributed Notes 21 notes

down

120

Jimmy ¶

13 years ago


Things to be aware of when using week numbers with years.

<?php
echo date("YW", strtotime("2011-01-07")); // gives 201101
echo date("YW", strtotime("2011-12-31")); // gives 201152
echo date("YW", strtotime("2011-01-01")); // gives 201152 too
?>

BUT

<?php
echo date("oW", strtotime("2011-01-07")); // gives 201101
echo date("oW", strtotime("2011-12-31")); // gives 201152
echo date("oW", strtotime("2011-01-01")); // gives 201052 (Year is different than previous example)
?>

Reason:
Y is year from the date
o is ISO-8601 year number
W is ISO-8601 week number of year

Conclusion:
if using 'W' for the week number use 'o' for the year.

down

ivijan dot stefan at gmail dot com ¶

9 years ago


If you have a problem with the different time zone, this is the solution for that.
<?php
// first line of PHP
$defaultTimeZone='UTC';
if(date_default_timezone_get()!=$defaultTimeZone)) date_default_timezone_set($defaultTimeZone);

// somewhere in the code
function _date($format="r", $timestamp=false, $timezone=false)
{
    $userTimezone = new DateTimeZone(!empty($timezone) ? $timezone : 'GMT');
    $gmtTimezone = new DateTimeZone('GMT');
    $myDateTime = new DateTime(($timestamp!=false?date("r",(int)$timestamp):date("r")), $gmtTimezone);
    $offset = $userTimezone->getOffset($myDateTime);
    return date($format, ($timestamp!=false?(int)$timestamp:$myDateTime->format('U')) + $offset);
}

/* Example */
echo 'System Date/Time: '.date("Y-m-d | h:i:sa").'<br>';
echo 'New York Date/Time: '._date("Y-m-d | h:i:sa", false, 'America/New_York').'<br>';
echo 'Belgrade Date/Time: '._date("Y-m-d | h:i:sa", false, 'Europe/Belgrade').'<br>';
echo 'Belgrade Date/Time: '._date("Y-m-d | h:i:sa", 514640700, 'Europe/Belgrade').'<br>';
?>
This is the best and fastest solution for this problem. Working almost identical to date() function only as a supplement has the time zone option.

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adityabhai at gmail dot com ¶

11 years ago


For Microseconds, we can get by following:

echo date('Ymd His'.substr((string)microtime(), 1, 8).' e');

Thought, it might be useful to someone !

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FiraSEO ¶

11 years ago


this how you make an HTML5 <time> tag correctly

<?php

echo '<time datetime="'.date('c').'">'.date('Y - m - d').'</time>';

?>

in the "datetime" attribute you should put a machine-readable value which represent time , the best value is a full time/date with ISO 8601 ( date('c') ) ,,, the attr will be hidden from users

and it doesn't really matter what you put as a shown value to the user,, any date/time format is okay !

This is very good for SEO especially search engines like Google .

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remindfwd at gmail dot com ¶

2 years ago


To get one month back, then simply write:

```
$oneMonthBack = date( DateTime::ISO8601, strtotime('-1 month' )); 
```

// results (for eg.)
// 2021-12-21T13:44:05+0000

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Bas Vijfwinkel ¶

12 years ago


Note that some formatting options are different from MySQL.
For example using a 24 hour notation without leading zeros is the option '%G' in PHP but '%k' in MySQL.
When using dynamically generated date formatting string, be careful to generate the correct options for either PHP or MySQL.

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SpikeDaCruz ¶

18 years ago


The following function will return the date (on the Gregorian calendar) for Orthodox Easter (Pascha).  Note that incorrect results will be returned for years less than 1601 or greater than 2399. This is because the Julian calendar (from which the Easter date is calculated) deviates from the Gregorian by one day for each century-year that is NOT a leap-year, i.e. the century is divisible by 4 but not by 10.  (In the old Julian reckoning, EVERY 4th year was a leap-year.)



This algorithm was first proposed by the mathematician/physicist Gauss.  Its complexity derives from the fact that the calculation is based on a combination of solar and lunar calendars.



<?php

function getOrthodoxEaster($date){

  /*

   Takes any Gregorian date and returns the Gregorian

   date of Orthodox Easter for that year.

  */

  $year = date("Y", $date);

  $r1 = $year % 19;

  $r2 = $year % 4;

  $r3 = $year % 7;

  $ra = 19 * $r1 + 16;

  $r4 = $ra % 30;

  $rb = 2 * $r2 + 4 * $r3 + 6 * $r4;

  $r5 = $rb % 7;

  $rc = $r4 + $r5;

  //Orthodox Easter for this year will fall $rc days after April 3

  return strtotime("3 April $year + $rc days");

}

?>

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Anonymous ¶

10 years ago


It's common for us to overthink the complexity of date/time calculations and underthink the power and flexibility of PHP's built-in functions.  Consider http://php.net/manual/en/function.date.php#108613

<?php
function get_time_string($seconds)
{
    return date('H:i:s', strtotime("2000-01-01 + $seconds SECONDS"));
}

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matthew dot hotchen at worldfirst dot com ¶

10 years ago


FYI: there's a list of constants with predefined formats on the DateTime object, for example instead of outputting ISO 8601 dates with:

<?php
echo date('c');
?>

or

<?php
echo date('Y-m-d\TH:i:sO');
?>

You can use

<?php
echo date(DateTime::ISO8601);
?>

instead, which is much easier to read.

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-1

bruslbn at gmail dot com ¶

6 years ago


In order to define leap year you must considre not only that year can be divide by 4!

The correct alghoritm is:

if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)

So the code should look like this:

if($year%4 == 0 && $year%100 != 0) {
    $leapYear = 1;
} elseif($year%400 == 0) {
    $leapYear = 1;                           
} else {
    $leapYear = 0;
}

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-2

ghotinet ¶

14 years ago


Most spreadsheet programs have a rather nice little built-in function called NETWORKDAYS to calculate the number of business days (i.e. Monday-Friday, excluding holidays) between any two given dates. I couldn't find a simple way to do that in PHP, so I threw this together. It replicates the functionality of OpenOffice's NETWORKDAYS function - you give it a start date, an end date, and an array of any holidays you want skipped, and it'll tell you the number of business days (inclusive of the start and end days!) between them.

I've tested it pretty strenuously but date arithmetic is complicated and there's always the possibility I missed something, so please feel free to check my math.

The function could certainly be made much more powerful, to allow you to set different days to be ignored (e.g. "skip all Fridays and Saturdays but include Sundays") or to set up dates that should always be skipped (e.g. "skip July 4th in any year, skip the first Monday in September in any year"). But that's a project for another time.

<?php

function networkdays($s, $e, $holidays = array()) {
    // If the start and end dates are given in the wrong order, flip them.    
    if ($s > $e)
        return networkdays($e, $s, $holidays);

    // Find the ISO-8601 day of the week for the two dates.
    $sd = date("N", $s);
    $ed = date("N", $e);

    // Find the number of weeks between the dates.
    $w = floor(($e - $s)/(86400*7));    # Divide the difference in the two times by seven days to get the number of weeks.
    if ($ed >= $sd) { $w--; }        # If the end date falls on the same day of the week or a later day of the week than the start date, subtract a week.

    // Calculate net working days.
    $nwd = max(6 - $sd, 0);    # If the start day is Saturday or Sunday, add zero, otherewise add six minus the weekday number.
    $nwd += min($ed, 5);    # If the end day is Saturday or Sunday, add five, otherwise add the weekday number.
    $nwd += $w * 5;        # Add five days for each week in between.

    // Iterate through the array of holidays. For each holiday between the start and end dates that isn't a Saturday or a Sunday, remove one day.
    foreach ($holidays as $h) {
        $h = strtotime($h);
        if ($h > $s && $h < $e && date("N", $h) < 6)
            $nwd--;
    }

    return $nwd;
}

$start = strtotime("1 January 2010");
$end = strtotime("13 December 2010");

// Add as many holidays as desired.
$holidays = array();
$holidays[] = "4 July 2010";            // Falls on a Sunday; doesn't affect count
$holidays[] = "6 September 2010";        // Falls on a Monday; reduces count by one

echo networkdays($start, $end, $holidays);    // Returns 246

?>

Or, if you just want to know how many work days there are in any given year, here's a quick function for that one:

<?php

function workdaysinyear($y) {
    $j1 = mktime(0,0,0,1,1,$y);
    if (date("L", $j1)) {
        if (date("N", $j1) == 6)
            return 260;
        elseif (date("N", $j1) == 5 or date("N", $j1) == 7)
            return 261;
        else
            return 262;
    }
    else {
        if (date("N", $j1) == 6 or date("N", $j1) == 7)
            return 260;
        else
            return 261;
    }
}

?>

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-6

Charlie ¶

9 years ago


For HTML5 datetime-local HTML input controls (http://www.w3.org/TR/html-markup/input.datetime-local.html) use format example: 1996-12-19T16:39:57

To generate this, escape the 'T', as shown below:

<?php
date('Y-m-d\TH:i:s');
?>

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-6

mirco dot babin at gmail dot com ¶

6 years ago


One important thing you should remember is that the timestamp value returned by time() is time-zone agnostic and gets the number of seconds since 1 January 1970 at 00:00:00 UTC. This means that at a particular point in time, this function will return the same value in the US, Europe, India, Japan, ...

date() will format a time-zone agnostic timestamp according to the default timezone set with date_default_timezone_set(...). Local time. If you want to output as UTC time use:

<?php
function dateUTC($format, $timestamp = null)
{
    if ($timestamp === null) $timestamp = time();

    $tz = date_default_timezone_get();
    date_default_timezone_set('UTC');

    $result = date($format, $timestamp);

    date_default_timezone_set($tz);
    return $result;
}
/>

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-9

david dot thomas at elliott-thomas dot com dot au ¶

8 years ago


Prior to PHP 5.6.23,  Relative Formats for the start of the week aligned with PHP's (0=Sunday,6=Saturday). Since 5.6.23,  Relative Formats for the start of the week align with ISO-8601 (1=Monday,7=Sunday). (http://php.net/manual/en/datetime.formats.relative.php)

This can produce different, and seemingly incorrect, results depending on your PHP version and your choice of 'w' or 'N' for the Numeric representation of the day of the week: 

<?php
echo "Today is Sun 2 Oct 2016, day ",date('w',strtotime('2016-10-02'))," of this week. ";  
echo "Day ",date('w',strtotime('2016-10-02 Monday next week'))," of next week is ",date('d M Y',strtotime('2016-10-02 Monday next week')),"<br />";

echo "Today is Sun 2 Oct 2016, day ",date('N',strtotime('2016-10-02'))," of this week. ";  
echo "Day ",date('w',strtotime('2016-10-02 Monday next week'))," of next week is ",date('d M Y',strtotime('2016-10-02 Monday next week'));
?> 

Prior to PHP 5.6.23, this results in: 

Today is Sun 2 Oct 2016, day 0 of this week. Day 1 of next week is 10 Oct 2016
Today is Sun 2 Oct 2016, day 7 of this week. Day 1 of next week is 10 Oct 2016

Since PHP 5.6.23, this results in: 

Today is Sun 2 Oct 2016, day 0 of this week. Day 1 of next week is 03 Oct 2016
Today is Sun 2 Oct 2016, day 7 of this week. Day 1 of next week is 03 Oct 2016

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-8

rc at macshot dot de ¶

7 years ago


At least in PHP 5.5.38 date('j.n.Y', 2222222222) gives a result of 2.6.2040.

So date is not longer limited to the minimum and maximum values for a 32-bit signed integer as timestamp.

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-10

Anonymous ¶

8 years ago


If timestamp is a string, date converts it to an integer in a possibly unexpected way:

<?php
echo (int)'0x10'; //0
echo intval('0x10'); //0
echo date('s', '0x10'); //gives 16
//however, no octal conversion:
echo date('s', '010'); //gives 10
?>

(PHP 5.6.16)

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-7

Jaap ¶

4 years ago


For users looking to format a unix timestamp with microseconds to mysql datetime, this function should do the trick:
<?php
        function sqlDateTimeFromMicroTimestamp(int $microtimestamp):string{
            $dt = new \DateTimeImmutable(); 
            $normalTimestamp = (int)floor($microtimestamp / 1000000);
            $sqlTimestampWithoutMicroseconds = $dt->setTimestamp($normalTimestamp)->format('Y-m-d H:i:s');
            $sqlTimestampWithMicroseconds = $sqlTimestampWithoutMicroseconds . '.'. ($microtimestamp % 1000000);
            return $sqlTimestampWithMicroseconds;
        }
?>

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-13

bruslbn at gmail dot com ¶

6 years ago


In order to define leap year you must considre not only that year can be divide by 4!

The correct alghoritm is:

if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)

So the code should look like this:

if($year%4 == 0 && $year%100 != 0) {
    $leapYear = 1;
} elseif($year%400 == 0) {
    $leapYear = 1;                           
} else {
    $leapYear = 0;
}

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-8

mparsa1372 at gmail dot com ¶

3 years ago


The example below formats today's date in three different ways:

<?php
echo "Today is " . date("Y/m/d") . "<br>";
echo "Today is " . date("Y.m.d") . "<br>";
echo "Today is " . date("Y-m-d") . "<br>";
echo "Today is " . date("l");
?>

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-21

bruslbn at gmail dot com ¶

6 years ago


In order to define leap year you must considre not only that year can be divide by 4!

The correct alghoritm is:

if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)

So the code should look like this:

if($year%4 == 0 && $year%100 != 0) {
    $leapYear = 1;
} elseif($year%400 == 0) {
    $leapYear = 1;                           
} else {
    $leapYear = 0;
}

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-28

arth dot inbox at gmail dot com ¶

6 years ago


Looks like date('u') is not microseconds, but is positive difference from rest part.

php > echo (DateTime::createFromFormat('U.u', '-128649659.999998'))->format('Y-m-d H:i:s.u U.u');
1965-12-03 23:59:01.999998 -128649659.999998

`U.u` parsed and formatted same, but means not 1965-12-03 23:59:00.000002.
Other words correct timestamp for example above is (-128649659 + 0.999998).  

Less confusing format for it is:  

php > echo DateTime::createFromFormat('U\+0.u', '-128649660+0.000002')->format('Y-m-d H:i:s.u');
1965-12-03 23:59:00.000002

Is that bug or feature?

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