PHP: date - Manual

有效的时间戳典型范围是格林威治时间 1901 年 12 月 13 日 20:45:54 到 2038 年 1 月 19 日 03:14:07。（此范围符合 32 位有符号整数的最小值和最大值）。不过在 PHP 5.1 之前此范围在某些系统（如 Windows）中限制为从 1970 年 1 月 1 日到 2038 年 1 月 19 日。要将字符串表达的时间转换成时间戳，应该使用 strtotime() 。此外一些数据库有一些函数将其时间格式转换成时间戳（例如 MySQL 的 » UNIX_TIMESTAMP 函数）。 <?php
// 设定要用的默认时区。自 PHP 5.1 可用
date_default_timezone_set ( 'UTC' );

// 输出类似：Monday
echo date ( "l" );

// 输出类似：Monday 15th of August 2005 03:12:46 PM
echo date ( 'l dS \of F Y h:i:s A' );

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

/* 在格式参数中使用常量 */
// 输出类似：Wed, 25 Sep 2013 15:28:57 -0700
echo date ( DATE_RFC2822 );

// 输出类似：2000-07-01T00:00:00+00:00
echo date ( DATE_ATOM , mktime ( 0 , 0 , 0 , 7 , 1 , 2000 ));
?> 一些使用 date() 格式化日期的例子。注意要转义所有其它的字符，因为目前有特殊含义的字符会产生不需要的结果，而其余字符在 PHP 将来的版本中可能会被用上。当转义时，注意用单引号以避免类似 \n 的字符变成了换行符。

示例 #4 date() 格式举例

<?php
// 假定今天是：March 10th, 2001, 5:16:18 pm
$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 z ' ); // 05-16-17, 10-03-01, 1631 1618 6 Fripm01
$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 15:16:08 MST 2001
$today = date ( 'H:m:s \m \i\s\ \m\o\n\t\h' ); // 17:03:17 m is month
$today = date ( "H:i:s" ); // 17:16:17
$today = date ( "Y-m-d H:i:s" ); // 2001-03-10 17:16:18 （MySQL DATETIME 格式）
?> ISO-8601 规范的年份，同

格式。有一种情况除外：当 ISO 的周数（

）属于前一年或者后一年时，会返回前一年或者后一年的年份数字表达。属于前一年或者后一年时，会返回前一年或者后一年的年份数字表达。（PHP 5.1.0 新加）示例：

或

4 位数字的年份示例：

或

2 位数字的年份示例：

或

上午还是下午，2 位小写字符

am

或

pm

上午还是下午，2 位大写字符

AM

或

PM

斯沃琪因特网时间从

到

小时，12时制，无前导 0 从

到

小时，24时制，无前导 0 从

到

小时，12时制，有前导 0 的 2 位数字从

到

小时，24时制，有前导 0 的 2 位数字

through

分钟，有前导 0 的 2 位数字从

到

秒，有前导 0 的 2 位数字从

到

毫秒（PHP 5.2.2 新加）示例：

时区标识（PHP 5.1.0 新加）示例:

UTC

GMT


   Atlantic/Azores

（大写字母 i）是否夏令时如果是夏令时则返回

，反之返回

。和格林威治时间（GMT）的时差，以小时为单位示例：


   +0200

和格林威治时间（GMT）的时差，包括小时和分钟，小时和分钟之间使用冒号（:）分隔（PHP 5.1.3 新加）示例：


   +02:00

示例：

EST

MDT

... 以秒为单位的时区偏移量。UTC 以西的时区返回负数，UTC 以东的时区返回正数。从


   -43200

到

完整的日期/时间 ISO 8601 日期及时间（PHP 5 新加） 2004-02-12T15:19:21+00:00 » RFC 2822 格式的日期和时间示例：


   Thu, 21 Dec 2000 16:01:07 +0200

自 1970 年 1 月 1 日 0 时 0 分 0 秒（GMT 时间）以来的时间，以秒为单位参见 time() <?php
// 设置默认时区。PHP 5.1 之后版本可用
date_default_timezone_set ( 'UTC' );

// 输出类似： Monday
echo date ( "l" );

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

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

/* 使用格式常量 */
// 输出类似： Mon, 15 Aug 2005 15:12:46 UTC
echo date ( DATE_RFC822 );

// 输出类似：2000-07-01T00:00:00+00:00
echo date ( DATE_ATOM , mktime ( 0 , 0 , 0 , 7 , 1 , 2000 ));
?> 需要注意的是，即使是对于当前来说并不具有特殊含义的字符，也要像对待具有特殊含义的字符那样进行转义，以避免函数返回非预期的值。因为可能在将来的 PHP 版本中，这些字符会被赋予特殊的含义。进行转义的时候，请确保使用单引号，以避免 \n 被解释为换行符号。

示例 #8 date() 函数格式化

<?php
// 假设今天是 2001 年 3 月 10 日下午 5 点 16 分 18 秒，
// 并且位于山区标准时间（MST）时区

$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
?>

strftime() - 根据区域设置格式化本地时间／日期

time() - 返回当前的 Unix 时间戳

strtotime() - 将任何字符串的日期时间描述解析为 Unix 时间戳

预定义的日期时间常量

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. 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" ). ' ' ;
echo 'New York Date/Time: ' . _date ( "Y-m-d | h:i:sa" , false , 'America/New_York' ). ' ' ;
echo 'Belgrade Date/Time: ' . _date ( "Y-m-d | h:i:sa" , false , 'Europe/Belgrade' ). ' ' ;
echo 'Belgrade Date/Time: ' . _date ( "Y-m-d | h:i:sa" , 514640700 , 'Europe/Belgrade' ). ' ' ;
?>
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. 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;
} 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 . 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" ));
} 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 ;
}
?> 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" );
}
?> 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 ;
}
}

?> 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 ;
}
/> 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' )), " " ;

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 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;
} 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;
} 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?