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.
date
(PHP 4, PHP 5, PHP 7, PHP 8)
date — 格式化 Unix 时间戳
说明
使用指定整数 timestamp（Unix 时间戳），或者使用当前时间（如果没有指定时间戳），返回相应的指定格式的格式化字符串。换句话说，timestamp
是可选的，默认值是 time()。
警告
Unix 时间戳不处理时区。使用 DateTimeImmutable 及其 DateTimeInterface::format() 格式化方法来格式化携带时区的日期/时间信息。
参数
-
format -
DateTimeInterface::format() 接受的格式。
注意: date() 将始终生成
000000作为微秒，因为它采用 int 参数，而如果 DateTime 是用微秒创建的，则 DateTime::format() 支持微秒。 -
timestamp -
可选的
timestamp参数是一个 int 的 Unix 时间戳，如未指定或是null，参数值默认为当前本地时间。也就是说，其值默认为 time() 的返回值。
返回值
返回格式化后的日期字符串。
错误／异常
在每次调用日期/时间函数时，如果时区无效则会引发 E_NOTICE 错误。参见
date_default_timezone_set()。
更新日志
| 版本 | 说明 |
|---|---|
| 8.0.0 |
现在 timestamp 允许为 null。
|
示例
示例 #1 date() 示例
<?php
// 设置要使用的默认时区。
date_default_timezone_set('UTC');
// 打印类似：Monday
echo date("l");
// 打印类似：Monday 8th of August 2005 03:12:46 PM
echo date('l jS \of F Y h:i:s A');
// 打印：July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", mktime(0, 0, 0, 7, 1, 2000));
/* 使用格式化参数中的常量 */
// 打印类似：Wed, 25 Sep 2013 15:28:57 -0700
echo date(DATE_RFC2822);
// 打印类似：2000-07-01T00:00:00+00:00
echo date(DATE_ATOM, mktime(0, 0, 0, 7, 1, 2000));
?>可以对格式化字符串中的解析字符，在其前面添加反斜线来防止解析。如果带有反斜线的字符已经是特殊字符，需要对反斜线进行再次转义。
示例 #2 date() 中的转义字符
<?php
// 打印类似： Wednesday the 15th
echo date('l \t\h\e jS');
?>可以一起使用 date() 和 mktime() 来查找过去或未来的日期时间。
示例 #3 date() 和 mktime() 示例
<?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);
?>注意:
由于存在夏令时时间，所以此方案相对于直接在时间戳上加/减秒数要更加可靠。
date() 格式化的一些示例。请注意，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，以避免产生非预期的值，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 这样的字符被解释为换行符号。
示例 #4 date() 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
$today = date("Y-m-d H:i:s"); // 2001-03-10 17:16:18（MySQL DATETIME 格式）
?>如果需要将日期时间格式化为其他语言，使用 IntlDateFormatter::format() 而不是 date()。
注释
注意:
使用 strtotime() 将字符串表示的日期转换为时间戳。另外，一些数据库产品也提供了将日期格式转换为时间戳的函数（例如 MySQL 中的 » UNIX_TIMESTAMP 函数）。
小技巧
请求的开始时间可以从变量 $_SERVER['REQUEST_TIME'] 中获取。
参见
- DateTimeImmutable::__construct() - Returns new DateTimeImmutable object
- DateTimeInterface::format() - 按照指定格式返回格式化后的日期
- gmdate() - 格式化 GMT/UTC 日期／时间
- idate() - 将本地日期/时间格式化为整数
- getdate() - 获取日期/时间信息
- getlastmod() - 获取页面最后修改的时间
- mktime() - 取得一个日期的 Unix 时间戳
- IntlDateFormatter::format() - Format the date/time value as a string
- time() - 返回当前的 Unix 时间戳
- 预定义 DateTime 常量
add a note
User Contributed Notes 21 notes
Jimmy ¶
12 years ago
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.
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 !
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 .
remindfwd at gmail dot com ¶
2 years ago
To get one month back, then simply write:
```
$oneMonthBack = date( DateTime::ISO8601, strtotime('-1 month' ));
```
// results (for eg.)
// 2021-12-21T13:44:05+0000
Bas Vijfwinkel ¶
12 years ago
Note that some formatting options are different from MySQL.
For example using a 24 hour notation without leading zeros is the option '%G' in PHP but '%k' in MySQL.
When using dynamically generated date formatting string, be careful to generate the correct options for either PHP or MySQL.
SpikeDaCruz ¶
18 years ago
The following function will return the date (on the Gregorian calendar) for Orthodox Easter (Pascha). Note that incorrect results will be returned for years less than 1601 or greater than 2399. This is because the Julian calendar (from which the Easter date is calculated) deviates from the Gregorian by one day for each century-year that is NOT a leap-year, i.e. the century is divisible by 4 but not by 10. (In the old Julian reckoning, EVERY 4th year was a leap-year.)
This algorithm was first proposed by the mathematician/physicist Gauss. Its complexity derives from the fact that the calculation is based on a combination of solar and lunar calendars.
<?php
function getOrthodoxEaster($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 + $rc days");
}
?>
Anonymous ¶
10 years ago
It's common for us to overthink the complexity of date/time calculations and underthink the power and flexibility of PHP's built-in functions. Consider http://php.net/manual/en/function.date.php#108613
<?php
function get_time_string($seconds)
{
return date('H:i:s', strtotime("2000-01-01 + $seconds SECONDS"));
}
matthew dot hotchen at worldfirst dot com ¶
10 years ago
FYI: there's a list of constants with predefined formats on the DateTime object, for example instead of outputting ISO 8601 dates with:
<?php
echo date('c');
?>
or
<?php
echo date('Y-m-d\TH:i:sO');
?>
You can use
<?php
echo date(DateTime::ISO8601);
?>
instead, which is much easier to read.
bruslbn at gmail dot com ¶
6 years ago
In order to define leap year you must considre not only that year can be divide by 4!
The correct alghoritm is:
if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)
So the code should look like this:
if($year%4 == 0 && $year%100 != 0) {
$leapYear = 1;
} elseif($year%400 == 0) {
$leapYear = 1;
} else {
$leapYear = 0;
}
ghotinet ¶
13 years ago
Most spreadsheet programs have a rather nice little built-in function called NETWORKDAYS to calculate the number of business days (i.e. Monday-Friday, excluding holidays) between any two given dates. I couldn't find a simple way to do that in PHP, so I threw this together. It replicates the functionality of OpenOffice's NETWORKDAYS function - you give it a start date, an end date, and an array of any holidays you want skipped, and it'll tell you the number of business days (inclusive of the start and end days!) between them.
I've tested it pretty strenuously but date arithmetic is complicated and there's always the possibility I missed something, so please feel free to check my math.
The function could certainly be made much more powerful, to allow you to set different days to be ignored (e.g. "skip all Fridays and Saturdays but include Sundays") or to set up dates that should always be skipped (e.g. "skip July 4th in any year, skip the first Monday in September in any year"). But that's a project for another time.
<?php
function networkdays($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", $j1) == 7)
return 260;
else
return 261;
}
}
?>
Charlie ¶
8 years ago
For HTML5 datetime-local HTML input controls (http://www.w3.org/TR/html-markup/input.datetime-local.html) use format example: 1996-12-19T16:39:57
To generate this, escape the 'T', as shown below:
<?php
date('Y-m-d\TH:i:s');
?>
mirco dot babin at gmail dot com ¶
5 years ago
One important thing you should remember is that the timestamp value returned by time() is time-zone agnostic and gets the number of seconds since 1 January 1970 at 00:00:00 UTC. This means that at a particular point in time, this function will return the same value in the US, Europe, India, Japan, ...
date() will format a time-zone agnostic timestamp according to the default timezone set with date_default_timezone_set(...). Local time. If you want to output as UTC time use:
<?php
function dateUTC($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;
}
/>
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
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.
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)
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 $sqlTimestampWithMicroseconds;
}
?>
bruslbn at gmail dot com ¶
6 years ago
In order to define leap year you must considre not only that year can be divide by 4!
The correct alghoritm is:
if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)
So the code should look like this:
if($year%4 == 0 && $year%100 != 0) {
$leapYear = 1;
} elseif($year%400 == 0) {
$leapYear = 1;
} else {
$leapYear = 0;
}
mparsa1372 at gmail dot com ¶
3 years ago
The example below formats today's date in three different ways:
<?php
echo "Today is " . date("Y/m/d") . "<br>";
echo "Today is " . date("Y.m.d") . "<br>";
echo "Today is " . date("Y-m-d") . "<br>";
echo "Today is " . date("l");
?>
bruslbn at gmail dot com ¶
6 years ago
In order to define leap year you must considre not only that year can be divide by 4!
The correct alghoritm is:
if (year is not divisible by 4) then (it is a common year)
else if (year is not divisible by 100) then (it is a leap year)
else if (year is not divisible by 400) then (it is a common year)
else (it is a leap year)
So the code should look like this:
if($year%4 == 0 && $year%100 != 0) {
$leapYear = 1;
} elseif($year%400 == 0) {
$leapYear = 1;
} else {
$leapYear = 0;
}
arth dot inbox at gmail dot com ¶
5 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?
