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date

(PHP 4, PHP 5, PHP 7)

dateローカルの日付/時刻を書式化する

説明

date ( string $format [, int $timestamp = time() ] ) : string

指定された引数 timestamp を、与えられた フォーマット文字列によりフォーマットし、日付文字列を返します。 タイムスタンプが与えられない場合は、現在の時刻が使われます。 つまり timestamp はオプションであり そのデフォルト値は time() の値です。

パラメータ

format

DateTimeInterface::format() が受け入れるフォーマット。

timestamp

オプションのパラメータ timestamp は、 int 型の Unix タイムスタンプです。 timestamp が指定されなかった場合のデフォルト値は、 現在の時刻です。言い換えると、デフォルトは time() の返り値となります。

返り値

日付を表す文字列を返します。 timestamp に数字以外が使用された場合は FALSE が返され、E_WARNING レベルのエラーが発生します。

エラー / 例外

すべての日付/時刻関数は、 有効なタイムゾーンが設定されていない場合に E_NOTICE を発生させます。また、システム設定のタイムゾーンあるいは環境変数 TZ を使用した場合には E_STRICT あるいは E_WARNING を発生させます。 date_default_timezone_set() も参照ください。

例1 date() の例

<?php
// 使用するデフォルトのタイムゾーンを指定します。PHP 5.1 以降で使用可能です。
date_default_timezone_set('UTC');


// 結果は、たとえば Monday のようになります。
echo date("l");

// 結果は、たとえば Monday 8th of August 2005 03:12:46 PM のようになります。
echo date('l jS \of F Y h:i:s A');

// 結果は July 1, 2000 is on a Saturday となります。
echo "July 1, 2000 is on a " date("l"mktime(000712000));

/* 書式指定パラメータに、定数を使用します。 */
// 結果は、たとえば Wed, 25 Sep 2013 15:28:57 -0700 のようになります。
echo date(DATE_RFC2822);

// 結果は、たとえば 2000-07-01T00:00:00+00:00 のようになります。
echo date(DATE_ATOMmktime(000712000));
?>

前にバックスラッシュを付けてエスケープすることにより、 フォーマット文字列として認識される文字が展開されることを防止することができます。 バックスラッシュ付きの文字は既に特別なシーケンスであり、 バックスラッシュもエスケープすることが必要となる可能性があります。

例2 date() の文字をエスケープする

<?php
// Wednesday the 15th のように出力
echo date('l \t\h\e jS');
?>

date()mktime() の両方を用いて、未来または過去の日付を知ることができます。

例3 date()mktime() の例

<?php
$tomorrow  
mktime(000date("m")  , date("d")+1date("Y"));
$lastmonth mktime(000date("m")-1date("d"),   date("Y"));
$nextyear  mktime(000date("m"),   date("d"),   date("Y")+1);
?>

注意:

サマータイムがあるため、日付や月の秒数を単純にタイムスタンプに 可減算するよりもより信頼性があります。

date() フォーマットのいくつかの例を示します。 現在の実装で特別な意味がある文字や今後の PHP のバージョンで意味が 割り付けられるであろう文字については、望ましくない結果を避けるために エスケープする必要があることに注意してください。エスケープを する際には、改行文字 \n のような文字を回避するために シングルクォートを使用してください。

例4 date() のフォーマット指定

<?php
// 今日は March 10th, 2001, 5:16:18 pm であり、
// またタイムゾーンは Mountain Standard Time (MST) であるものとします

$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 (MySQL の DATETIME フォーマット)
?>

他の言語で日付をフォーマットするためには、date() のかわりに setlocale() および strftime() 関数を使用する必要があります。

注意

注意:

日付の文字列表現からタイムスタンプを生成するには、 strtotime() が使用できるでしょう。 さらに、いくつかのデータベースは(MySQL の » UNIX_TIMESTAMP 関数の ような)日付フォーマットからタイムスタンプへの変換関数を有しています。

ヒント

PHP 5.1 以降、$_SERVER['REQUEST_TIME'] によってリクエスト開始時のタイムスタンプが取得できるようになりました。

参考

  • gmdate() - GMT/UTC の日付/時刻を書式化する
  • idate() - ローカルな時刻/日付を整数として整形する
  • getdate() - 日付/時刻情報を取得する
  • getlastmod() - 最終更新時刻を取得する
  • mktime() - 日付を Unix のタイムスタンプとして取得する
  • strftime() - ロケールの設定に基づいてローカルな日付・時間をフォーマットする
  • time() - 現在の Unix タイムスタンプを返す
  • DateTimeImmutable::__construct() - 新しい DateTimeImmutable オブジェクトを返す
  • 定義済みの定数

add a note add a note

User Contributed Notes 19 notes

up
106
Jimmy
8 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.
up
12
matthew dot hotchen at worldfirst dot com
6 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.
up
13
adityabhai at gmail dot com
7 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 !
up
7
david dot thomas at elliott-thomas dot com dot au
4 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
up
4
bruslbn at gmail dot com
2 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;
}
up
11
ivijan dot stefan at gmail dot com
5 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.
up
9
Anonymous
6 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"));
}
up
11
FiraSEO
7 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 .
up
5
Charlie
4 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');
?>
up
4
Anonymous
4 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)
up
1
rc at macshot dot de
3 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.
up
0
Jaap
2 months 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;
        }
?>
up
3
SpikeDaCruz
14 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");
}
?>
up
1
Bas Vijfwinkel
8 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.
up
0
ghotinet
9 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;
    }
}

?>
up
-2
mirco dot babin at gmail dot com
1 year 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;
}
/>
up
-9
bruslbn at gmail dot com
2 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;
}
up
-11
arth dot inbox at gmail dot com
1 year 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?
up
-12
bruslbn at gmail dot com
2 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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