This dateDiff() function can take in just about any timestamp, including UNIX timestamps and anything that is accepted by strtotime(). It returns an array with the ability to split the result a couple different ways. I built this function to suffice any datediff needs I had. Hope it helps others too.
<?php
/********* dateDiff() function **********
* returns Array of Int values for difference between two dates
* $date1 > $date2 --> positive integers are returned
* $date1 < $date2 --> negative integers are returned
*
* $split recognizes the following:
* 'yw' = splits up years, weeks and days (default)
* 'y' = splits up years and days
* 'w' = splits up weeks and days
* 'd' = total days
*
* examples:
* $dif1 = dateDiff() or dateDiff('yw')
* $dif2 = dateDiff('y')
* $dif3 = dateDiff('w')
* $dif4 = dateDiff('d')
*
* assuming dateDiff returned 853 days, the above
* examples would have a print_r output of:
* $dif1 == Array( [y] => 2 [w] => 17 [d] => 4 )
* $dif2 == Array( [y] => 2 [d] => 123 )
* $dif3 == Array( [w] => 121 [d] => 6 )
* $dif4 == Array( [d] => 847 )
*
* note: [h] (hours), [m] (minutes), [s] (seconds) are always returned as elements of the Array
*/
function dateDiff($dt1, $dt2, $split='yw') {
$date1 = (strtotime($dt1) != -1) ? strtotime($dt1) : $dt1;
$date2 = (strtotime($dt2) != -1) ? strtotime($dt2) : $dt2;
$dtDiff = $date1 - $date2;
$totalDays = intval($dtDiff/(24*60*60));
$totalSecs = $dtDiff-($totalDays*24*60*60);
$dif['h'] = $h = intval($totalSecs/(60*60));
$dif['m'] = $m = intval(($totalSecs-($h*60*60))/60);
$dif['s'] = $totalSecs-($h*60*60)-($m*60);
// set up array as necessary
switch($split) {
case 'yw': # split years-weeks-days
$dif['y'] = $y = intval($totalDays/365);
$dif['w'] = $w = intval(($totalDays-($y*365))/7);
$dif['d'] = $totalDays-($y*365)-($w*7);
break;
case 'y': # split years-days
$dif['y'] = $y = intval($totalDays/365);
$dif['d'] = $totalDays-($y*365);
break;
case 'w': # split weeks-days
$dif['w'] = $w = intval($totalDays/7);
$dif['d'] = $totalDays-($w*7);
break;
case 'd': # don't split -- total days
$dif['d'] = $totalDays;
break;
default:
die("Error in dateDiff(). Unrecognized \$split parameter. Valid values are 'yw', 'y', 'w', 'd'. Default is 'yw'.");
}
return $dif;
}
?>
PHP - Manual: Date/Time 函数
2024-12-20
Date/Time 函数
目录
- checkdate — 验证一个格里高里日期
- date_add — 别名 DateTime::add
- date_create_from_format — 别名 DateTime::createFromFormat
- date_create_immutable_from_format — 别名 DateTimeImmutable::createFromFormat
- date_create_immutable — 别名 DateTimeImmutable::__construct
- date_create — create a new DateTime object
- date_date_set — 别名 DateTime::setDate
- date_default_timezone_get — 取得一个脚本中所有日期时间函数所使用的默认时区
- date_default_timezone_set — 设定用于一个脚本中所有日期时间函数的默认时区
- date_diff — 别名 DateTime::diff
- date_format — 别名 DateTime::format
- date_get_last_errors — 别名 DateTime::getLastErrors
- date_interval_create_from_date_string — 别名 DateInterval::createFromDateString
- date_interval_format — 别名 DateInterval::format
- date_isodate_set — 别名 DateTime::setISODate
- date_modify — 别名 DateTime::modify
- date_offset_get — 别名 DateTime::getOffset
- date_parse_from_format — Get info about given date formatted according to the specified format
- date_parse — 返回指定日期/时间的详细信息的关联数组
- date_sub — 别名 DateTime::sub
- date_sun_info — Returns an array with information about sunset/sunrise and twilight begin/end
- date_sunrise — 返回给定的日期与地点的日出时间
- date_sunset — 返回给定的日期与地点的日落时间
- date_time_set — 别名 DateTime::setTime
- date_timestamp_get — 别名 DateTime::getTimestamp
- date_timestamp_set — 别名 DateTime::setTimestamp
- date_timezone_get — 别名 DateTime::getTimezone
- date_timezone_set — 别名 DateTime::setTimezone
- date — 格式化一个本地时间／日期
- getdate — 取得日期／时间信息
- gettimeofday — 取得当前时间
- gmdate — 格式化一个 GMT/UTC 日期／时间
- gmmktime — 取得 GMT 日期的 UNIX 时间戳
- gmstrftime — 根据区域设置格式化 GMT/UTC 时间／日期
- idate — 将本地时间日期格式化为整数
- localtime — 取得本地时间
- microtime — 返回当前 Unix 时间戳和微秒数
- mktime — 取得一个日期的 Unix 时间戳
- strftime — 根据区域设置格式化本地时间／日期
- strptime — 解析由 strftime 生成的日期／时间
- strtotime — 将任何字符串的日期时间描述解析为 Unix 时间戳
- time — 返回当前的 Unix 时间戳
- timezone_abbreviations_list — 别名 DateTimeZone::listAbbreviations
- timezone_identifiers_list — 别名 DateTimeZone::listIdentifiers
- timezone_location_get — 别名 DateTimeZone::getLocation
- timezone_name_from_abbr — Returns the timezone name from abbreviation
- timezone_name_get — 别名 DateTimeZone::getName
- timezone_offset_get — 别名 DateTimeZone::getOffset
- timezone_open — 别名 DateTimeZone::__construct
- timezone_transitions_get — 别名 DateTimeZone::getTransitions
- timezone_version_get — 获取 timezonedb 的版本
add a note
User Contributed Notes 25 notes
daniel at globalnetstudios dot com ¶
17 years ago
mail at completeideas dot com ¶
17 years ago
For those who are using pre MYSQL 4.1.1, you can use:
TO_DAYS([Date Value 1])-TO_DAYS([Date Value 2])
For the same result as:
DATEDIFF([Date Value 1],[Date Value 2])
mincklerstraat at softhome dot net ¶
17 years ago
Before you get too advanced using date functions, be sure also to see the calendar functions at http://www.php.net/manual/en/ref.calendar.php .
bgold at matrix-consultants dot com ¶
15 years ago
When debugging code that stores date/time values in a database, you may find yourself wanting to know the date/time that corresponds to a given unix timestamp, or the timestamp for a given date & time.
The following script will do the conversion either way. If you give it a numeric timestamp, it will display the corresponding date and time. If you give it a date and time (in almost any standard format), it will display the timestamp.
All conversions are done for your locale/time zone.
<?php
while (true)
{
// Read a line from standard in.
echo "enter time to convert: ";
$inline = fgets(STDIN);
$inline = trim($inline);
if ($inline == "" || $inline == ".")
break;
// See if the line is a date.
$pos = strpos($inline, "/");
if ($pos === false) {
// not a date, should be an integer.
$date = date("m/d/Y G:i:s", $inline);
echo "int2date: $inline -> $date\n";
} else {
$itime = strtotime($inline);
echo "date2int: $inline -> $itime\n";
}
}
?>
nickaubert at america's biggest isp dot com ¶
18 years ago
I ran into an issue using a function that loops through an array of dates where the keys to the array are the Unix timestamp for midnight for each date. The loop starts at the first timestamp, then incremented by adding 86400 seconds (ie. 60 x 60 x 24). However, Daylight Saving Time threw off the accuracy of this loop, since certain days have a duration other than 86400 seconds. I worked around it by adding a couple of lines to force the timestamp to midnight at each interval.
<?php
$ONE_DAY = 90000; // can't use 86400 because some days have one hour more or less
for ( $each_timestamp = $start_time ; $each_timestamp <= $end_time ; $each_timestamp += $ONE_DAY) {
/* force midnight to compensate for daylight saving time */
$this_timestamp_array = getdate( $each_timestamp );
$each_timestamp = mktime ( 0 , 0 , 0 , $this_timestamp_array[mon] , $this_timestamp_array[mday] , $this_timestamp_array[year] );
// do some stuff...
}
?>
brighn (a) yahoo (.) com ¶
19 years ago
I needed a function that determined the last Sunday of the month. Since it's made for the website's "next meeting" announcement, it goes based on the system clock; also, if today is between Sunday and the end of the month, it figures out the last Sunday of *next* month. lastsunday() takes no arguments and returns the date as a string in the form "January 26, 2003". I could probably have streamlined this quite a bit, but at least it's transparent code. =)
<?php
/* The two functions calculate when the next meeting will
be, based on the assumption that the meeting will be on
the last Sunday of the month. */
function getlast($mon, $year) {
$daysinmonth = array(31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
$days = $daysinmonth[$mon-1];
if ($mon == 2 && ($year % 4) == 0 && (($year % 100) != 0 ||
($year % 400) == 0)) $days++;
if ($mon == 2 && ($year % 4) == 0 && ($year % 1000) != 0) $days++;
$lastday = getdate(mktime(0,0,0,$mon,$days,$year));
$wday = $lastday['wday'];
return getdate(mktime(0,0,0,$mon,$days-$wday,$year));
}
function lastsunday() {
$today = getdate();
$mon = $today['mon'];
$year = $today['year'];
$mday = $today['mday'];
$lastsun = getlast($mon, $year);
$sunday = $lastsun['mday'];
if ($sunday < $mday) {
$mon++;
if ($mon = 13) {
$mon = 1;
$year++;
}
$lastsun = getlast($mon, $year);
$sunday = $lastsun['mday'];
}
$nextmeeting = getdate(mktime(0,0,0,$mon,$sunday,$year));
$month = $nextmeeting['month'];
$mday = $nextmeeting['mday'];
$year = $nextmeeting['year'];
return "$month $mday, $year";
}
?>
php-contrib at i-ps dot nospam dot net ¶
20 years ago
Someone may find this info of some use:
Rules for calculating a leap year:
1) If the year divides by 4, it is a leap year (1988, 1992, 1996 are leap years)
2) Unless it divides by 100, in which case it isn't (1900 divides by 4, but was not a leap year)
3) Unless it divides by 400, in which case it is actually a leap year afterall (So 2000 was a leap year).
In practical terms, to work out the number of days in X years, multiply X by 365.2425, rounding DOWN to the last whole number, should give you the number of days.
The result will never be more than one whole day inaccurate, as opposed to multiplying by 365, which, over more years, will create a larger and larger deficit.
nightowl at NOS-PA-M dot uk2 dot net ¶
19 years ago
I wanted to find all records in my database which match the current week (for a call-back function). I made up this function to find the start and end of the current week :
<?php
function week($curtime) {
$date_array = getdate (time());
$numdays = $date_array["wday"];
$startdate = date("Y-m-d", time() - ($numdays * 24*60*60));
$enddate = date("Y-m-d", time() + ((7 - $numdays) * 24*60*60));
$week['start'] = $startdate;
$week['end'] = $enddate;
return $week;
}
?>
garyc at earthling dot net ¶
19 years ago
I needed to calculate the week number from a given date and vice versa, where the week starts with a Monday and the first week of a year may begin the year before, if the year begins in the middle of the week (Tue-Sun). This is the way weekly magazines calculate their issue numbers.
Here are two functions that do exactly that:
Hope somebody finds this useful.
Gary
<?php
/* w e e k n u m b e r -------------------------------------- //
weeknumber returns a week number from a given date (>1970, <2030)
Wed, 2003-01-01 is in week 1
Mon, 2003-01-06 is in week 2
Wed, 2003-12-31 is in week 53, next years first week
Be careful, there are years with 53 weeks.
// ------------------------------------------------------------ */
function weeknumber ($y, $m, $d) {
$wn = strftime("%W",mktime(0,0,0,$m,$d,$y));
$wn += 0; # wn might be a string value
$firstdayofyear = getdate(mktime(0,0,0,1,1,$y));
if ($firstdayofyear["wday"] != 1) # if 1/1 is not a Monday, add 1
$wn += 1;
return ($wn);
} # function weeknumber
/* d a t e f r o m w e e k ---------------------------------- //
From a weeknumber, calculates the corresponding date
Input: Year, weeknumber and day offset
Output: Exact date in an associative (named) array
2003, 12, 0: 2003-03-17 (a Monday)
1995, 53, 2: 1995-12-xx
...
// ------------------------------------------------------------ */
function datefromweek ($y, $w, $o) {
$days = ($w - 1) * 7 + $o;
$firstdayofyear = getdate(mktime(0,0,0,1,1,$y));
if ($firstdayofyear["wday"] == 0) $firstdayofyear["wday"] += 7;
# in getdate, Sunday is 0 instead of 7
$firstmonday = getdate(mktime(0,0,0,1,1-$firstdayofyear["wday"]+1,$y));
$calcdate = getdate(mktime(0,0,0,$firstmonday["mon"], $firstmonday["mday"]+$days,$firstmonday["year"]));
$date["year"] = $calcdate["year"];
$date["month"] = $calcdate["mon"];
$date["day"] = $calcdate["mday"];
return ($date);
} # function datefromweek
?>
th at definitynet dot com ¶
21 years ago
I had some problems with dates between mySQL and PHP. PHP had all these great date functions but I wanted to store a usable value in my database tables. In this case I was using TIMESTAMP(14) <or 'YYYYMMDDHHMMSS'>.
This is perhaps the easiest way I have found to pull the PHP usable UNIX Datestamp from my mySQL datestamp stored in the tables:
Use the mySQL UNIX_TIMESTAMP() function in your SQL definition string. i.e.
$sql= "SELECT field1, field2, UNIX_TIMESTAMP(field3) as your_date
FROM your_table
WHERE field1 = '$value'";
The query will return a temp table with coulms "field1" "Field2" "your_date"
The "your_date" will be formatted in a UNIX TIMESTAMP! Now you can use the PHP date() function to spew out nice date formats.
Sample using above $sql:
20010111002747 = Date Stored on mySQL table (TIMESTAMP(14))
979172867 = value returned as your_date in sql stmt (UNIX_TIMESTAMP)
if we use $newdate = date("F jS, Y -- g:ia", $row["your_date"]);
--(after fetching our array from the sql results of course)--
echo "$newdate"; --Will produce:
January 11th, 2001 -- 12:27am
Hope this helps someone out there!
koch.ro ¶
14 years ago
Not really elegant, but tells you, if your installed timezonedb is the most recent:
<?php
class TestDateTimeTimezonedbVersion extends PHPUnit_Framework_TestCase
{
public function testTimezonedbIsMostRecent()
{
ini_set( 'date.timezone', 'Europe/Berlin' );
ob_start();
phpinfo(INFO_MODULES);
$info = ob_get_contents();
ob_end_clean();
$start = strpos( $info, 'Timezone Database Version' ) + 29;
$this->assertTrue( FALSE !== $start, 'Seems there is no timezone DB installed' );
$end = strpos( $info, "\n", $start );
$installedVersion = substr( $info, $start, $end - $start );
exec( 'pecl remote-info timezonedb', &$output );
$availableVersion = substr( $output[2], 12 );
$this->assertEquals( $availableVersion, $installedVersion,
'The installed timezonedb is not actual. Installed: '.$installedVersion
.' available: '.$availableVersion
);
}
}
?>
Hans ¶
10 years ago
I needed a way to display an announcement on a shopping site, that would warn users that orders placed between a certain date range, would not be shipped until after a certain date.
I created this simple date detection code to display the notice on certain pages. You can just copy the code and save it to a file on the site and include it anywhere you need to perform a function, or display a notice.
<?php
/*
Code to show a message only for a certain time frame.
This is a simple include file that can be used to display a message
on any pages that use it.
Simply us a standard include instruction to this file on the page/s where
you want the notice to appear.
Written by Hans Kiesouw - hans at wotworx dot com
*/
$start = new DateTime('30-07-2011'); // DD-MM-YYYY
$endDate= new DateTime('07-08-2011'); // DD-MM-YYYY
$curdate = new DateTime(date('d-m-Y'));
if ($start <= $curdate && $curdate <= $endDate) {
/*
The message below will appear if the current date is between the start and
endDate - used standards HTML to ensure that any code will not be
escaped by PHP. You can use any code here to wish to execute for the
date range.
*/
?>
<p><strong><font color="#FF0000">Please Note: Any orders placed between July 30th and August 7th will only be shipped on August 8th. We apologize for any inconvenience and thank you for your order.</font></strong></p>
<?php
// don't forget this last bit, it ends the if statement!
}
?>
andreencinas at yahoo dot com dot br ¶
16 years ago
<?php
//function like dateDiff Microsoft
//not error in year Bissesto
function dateDiff($interval,$dateTimeBegin,$dateTimeEnd) {
//Parse about any English textual datetime
//$dateTimeBegin, $dateTimeEnd
$dateTimeBegin=strtotime($dateTimeBegin);
if($dateTimeBegin === -1) {
return("..begin date Invalid");
}
$dateTimeEnd=strtotime($dateTimeEnd);
if($dateTimeEnd === -1) {
return("..end date Invalid");
}
$dif=$dateTimeEnd - $dateTimeBegin;
switch($interval) {
case "s"://seconds
return($dif);
case "n"://minutes
return(floor($dif/60)); //60s=1m
case "h"://hours
return(floor($dif/3600)); //3600s=1h
case "d"://days
return(floor($dif/86400)); //86400s=1d
case "ww"://Week
return(floor($dif/604800)); //604800s=1week=1semana
case "m": //similar result "m" dateDiff Microsoft
$monthBegin=(date("Y",$dateTimeBegin)*12)+
date("n",$dateTimeBegin);
$monthEnd=(date("Y",$dateTimeEnd)*12)+
date("n",$dateTimeEnd);
$monthDiff=$monthEnd-$monthBegin;
return($monthDiff);
case "yyyy": //similar result "yyyy" dateDiff Microsoft
return(date("Y",$dateTimeEnd) - date("Y",$dateTimeBegin));
default:
return(floor($dif/86400)); //86400s=1d
}
}
?>
venoel at rin dot ru ¶
14 years ago
May be useful for somebody. This function takes on daylight saving time
<?php
Function DateDiff($date1,$date2) {
$timedifference=$date2-$date1;
$corr=date("I",$date2)-date("I",$date1);
$timedifference+=$corr;
return $timedifference;
}
?>
Example:
<?php
$d1=mktime(2,0,0,10,28,2007);
$d2=mktime(4,0,0,10,28,2007);
$period=DateDiff($d1,$d2);
printf("<br>%s",date("I d.m.Y H:i",$d1));
printf("<br>%u hour",$period/3600);
printf("<br>%s",date("I d.m.Y H:i",$d2));
?>
Getting 2 hour instead 3.
Darren Edwards ¶
14 years ago
I was looking for a solution where I could return the number of days, hours, Minutes and seconds between two entries in a table.
DATE_DIFF is not running on my mysql server as my provider uses mysql version 4.0.25
Solution was to use to days and std time functions to calculate the difference in one call.
The fields stored in the table(report_table) are
time(00:00:00),
date(0000-00-00) and record(enum) which tells the app the type of log stored. EG start or end of a report.
SELECT
(TO_DAYS( `end`.`date` ) - TO_DAYS( `start`.`date` ))
-
( second( `end`.`time` ) + (minute( `end`.`time` )*60) + (hour( `end`.`time` )*3600)
<
second( `start`.`time` ) + (minute( `start`.`time` )*60) + (hour( `start`.`time` )*3600))
AS `days` ,
SEC_TO_TIME(
(second( `end`.`time` ) + (minute( `end`.`time` )*60) + (hour( `end`.`time` )*3600) )
-
(second( `start`.`time` ) + (minute( `start`.`time` )*60) + (hour( `start`.`time` )*3600) )
) AS `hms`,
`start`.`time` as `start`,
`end`.`time` as `end`
FROM `report_table` AS `start` , `report_table` AS `end`
AND `start`.`record` = 'Report Begin'
AND `end`.`record` = 'Report End'
LIMIT 1
If there is no end of report then it will not return a result, as you would expect.
luck dot lil dot leprechaun at gmail dot com ¶
13 years ago
This is an easily extendable and pretty way to output human-readable date differences such as "1 day 2 hours ago", "6 months ago", "3 years 7 months 14 days 1 hour 4 minutes 16 seconds" etc etc.
Change "$levels = 2;" to whatever you want. A value of 1 will limit to only one number in the result ("3 days ago"). A value of 3 would result in up to three ("3 days 1 hour 2 minutes ago")
It can be used in the following ways:
echo compare_dates($start_date,$end_date);
echo compare_dates($end_date,$start_date);
echo compare_dates($start_date); //end date will be assumed as time();
<?php
function compare_dates($date1, $date2 = time())
{
$blocks = array(
array('name'=>'year','amount' => 60*60*24*365 ),
array('name'=>'month','amount' => 60*60*24*31 ),
array('name'=>'week','amount' => 60*60*24*7 ),
array('name'=>'day','amount' => 60*60*24 ),
array('name'=>'hour','amount' => 60*60 ),
array('name'=>'minute','amount' => 60 ),
array('name'=>'second','amount' => 1 )
);
$diff = abs($date1-$date2);
$levels = 2;
$current_level = 1;
$result = array();
foreach($blocks as $block)
{
if ($current_level > $levels) {break;}
if ($diff/$block['amount'] >= 1)
{
$amount = floor($diff/$block['amount']);
if ($amount>1) {$plural='s';} else {$plural='';}
$result[] = $amount.' '.$block['name'].$plural;
$diff -= $amount*$block['amount'];
$current_level++;
}
}
return implode(' ',$result).' ago';
}
?>
[EDIT BY danbrown AT php DOT net: Contains a bugfix supplied by (jorge AT dontspam DOT com) on 22-OCT-2009.]
php at elmegil dot net ¶
18 years ago
A much easier way to do days diff is to use Julian Days from the Calendar functions:
$start = gregoriantojd($smon, $sday, $syear);
$end = gregoriantojd($emon, $eday, $eyear);
$daysdiff = $end - $start;
You can see the obvious ways to wrap a function around that.
sagar ¶
14 years ago
<?php
####################################
# Provide week number and get start_timestamp and end_timestamp
#####################################
// this week number will come from the timeshare form
$week = 51;
$times = get_start_and_end_date_from_week($week);
$start_time = $times['start_timestamp'];
$end_time = $times['end_timestamp'];
function get_start_and_end_date_from_week ($w)
{
$y = date("Y", time());
$o = 6; // week starts from sunday by default
$days = ($w - 1) * 7 + $o;
$firstdayofyear = getdate(mktime(0,0,0,1,1,$y));
if ($firstdayofyear["wday"] == 0) $firstdayofyear["wday"] += 7;
# in getdate, Sunday is 0 instead of 7
$firstmonday = getdate(mktime(0,0,0,1,1-$firstdayofyear["wday"]+1,$y));
$calcdate = getdate(mktime(0,0,0,$firstmonday["mon"], $firstmonday["mday"]+$days,$firstmonday["year"]));
$sday = $calcdate["mday"];
$smonth = $calcdate["mon"];
$syear = $calcdate["year"];
$timestamp['start_timestamp'] = mktime(0, 0, 0, $smonth, $sday, $syear);
$timestamp['end_timestamp'] = $timestamp['start_timestamp'] + (60*60*24*7);
return $timestamp;
} # function datefromweek
?>
Leopoldo A dot Oducado (poducado at comfsm dot fm) ¶
15 years ago
Here is my function to count the number days, weeks, months, and year. I tried it below 1970 and it works.
<?php
function datecal($date,$return_value)
{
$date = explode("/", $date);
$month_begin = $date[0];
$month_begin_date = $date[1];
$year1 = $date[2];
$month_end = date("n");
$month_end_date = date("j");
$year2 = date("Y");
$days_old = 0;
$years_old = 0;
$months_old = 0;
if($month_begin==12)
{
$month = 1;
$year = $year1+1;
}
else
{
$month = $month_begin+1;
$year = $year1;
}
$begin_plus_days = cal_days_in_month(CAL_GREGORIAN, $month_begin, $year1) - $month_begin_date;
$end_minus_days = cal_days_in_month(CAL_GREGORIAN, $month_end, $year2) - $month_end_date;
while ($year <= $year2)
{
if($year == $year2)
{
$days_old = $days_old + cal_days_in_month(CAL_GREGORIAN, $month, $year);
if($month < $month_end)
{
$months_old = $months_old + 1;
$month = $month + 1;
}
elseif ($month==$month_end and $month_end_date >= $month_begin_date)
{
$year = $year2+1;
}
else
{
$year = $year2+1;
}
}
else
{
$days_old = $days_old + cal_days_in_month(CAL_GREGORIAN, $month, $year);
if ($month <= 11)
{
$month = $month + 1;
$months_old = $months_old + 1;
}
else
{
$month = 1;
$year = $year + 1;
$months_old = $months_old + 1;
}
}
}
$days_old = ($days_old + $begin_plus_days) - $end_minus_days;
if($return_value == "d")
{ return $days_old; }
elseif ($return_value == "w")
{ return intval($days_old/7); }
elseif ($return_value == "m")
{ return $months_old; }
elseif ($return_value == "y")
{ return intval($months_old/12); }
}
echo datecal("08/13/1975","m");
?>
glashio at xs4all dot nl ¶
16 years ago
Calculate Sum BusinessDays (Mon till Fri) between two date's :
<?php
function businessdays($begin, $end) {
$rbegin = is_string($begin) ? strtotime(strval($begin)) : $begin;
$rend = is_string($end) ? strtotime(strval($end)) : $end;
if ($rbegin < 0 || $rend < 0)
return 0;
$begin = workday($rbegin, TRUE);
$end = workday($rend, FALSE);
if ($end < $begin) {
$end = $begin;
$begin = $end;
}
$difftime = $end - $begin;
$diffdays = floor($difftime / (24 * 60 * 60)) + 1;
if ($diffdays < 7) {
$abegin = getdate($rbegin);
$aend = getdate($rend);
if ($diffdays == 1 && ($astart['wday'] == 0 || $astart['wday'] == 6) && ($aend['wday'] == 0 || $aend['wday'] == 6))
return 0;
$abegin = getdate($begin);
$aend = getdate($end);
$weekends = ($aend['wday'] < $abegin['wday']) ? 1 : 0;
} else
$weekends = floor($diffdays / 7);
return $diffdays - ($weekends * 2);
}
function workday($date, $begindate = TRUE) {
$adate = getdate($date);
$day = 24 * 60 * 60;
if ($adate['wday'] == 0) // Sunday
$date += $begindate ? $day : -($day * 2);
elseif ($adate['wday'] == 6) // Saterday
$date += $begindate ? $day * 2 : -$day;
return $date;
}
?>
Robb_Bean at gmx dot nospam dot net ¶
14 years ago
With PHP 5.1 and 5.2 the languages datetime support has changed. Although these functions should guess your local timezone settings, they may fail if using a default configuration in a "pre-5.1 way", which means setting no timezone for PHP. In the case PHP could not get a timezone it emits a E_STRICT warning. Note that this affects _all_ datetime functions and keep it in mind when porting software from previous versions to 5.1 or later! It may also confuse your error handling (this is the way I noticed that things have changed, since these changes are not documentated _here_).
References:
http://www.php.net/manual/de/migration51.datetime.php
http://www.php.net/manual/de/migration52.datetime.php
php at sarge dot ch ¶
18 years ago
Additional thisone here (didn't test it yet but should work :D):
<?php
/**
* Calculates the Difference between two timestamps
*
* @param integer $start_timestamp
* @param integer $end_timestamp
* @param integer $unit (default 0)
* @return string
* @access public
*/
function dateDifference($start_timestamp,$end_timestamp,$unit= 0){
$days_seconds_star= (23 * 56 * 60) + 4.091; // Star Day
$days_seconds_sun= 24 * 60 * 60; // Sun Day
$difference_seconds= $end_timestamp - $start_timestamp;
switch($unit){
case 3: // Days
$difference_days= round(($difference_seconds / $days_seconds_sun),2);
return 'approx. '.$difference_hours.' Days';
case 2: // Hours
$difference_hours= round(($difference_seconds / 3600),2);
return 'approx. '.$difference_hours.' Hours';
break;
case 1: // Minutes
$difference_minutes= round(($difference_seconds / 60),2);
return 'approx. '.$difference_minutes.' Minutes';
break;
default: // Seconds
if($difference_seconds > 1){
return $difference_seconds.' Seconds';
}
else{
return $difference_seconds.' Second';
}
}
}
?>
paez903 at hotmail dot com ¶
4 years ago
I do not have much programming in php and I hope I can help those that I want to do is that when entering in the form the date 1 and the date2 I calculate if between those two dates if they have passed 5 or more years and I add 3 more days taking As reference date 2, I do not know if I understand.
To see a theoretical example date = 10/01/2012 date2 = 23/07/2017
Between these two dates have passed 5 years, 6 months, and 13 days elapsed
Knowing this my serious conditional
If they are equal or more than 5 years but less than 10 years will be added 3 days
If they are equal or more than 10 years but less than 15 years will be added 6 days
If they are equal or more than 15 years but less than 20 years will be added 9 days
If they are equal to or more than 20 years will be added 12 days
Then having the conditionals
For this example would be case as it is greater than 5 years the result that I should show taking in days like date of departure the date2 = 23/07/2017 and to this date it is added 30 days that would be a constant and depending on the years like Is the example happens 5 years would be: date2 = 23/07/2017 + 30 days = 08/30/2017 + 3 days
= End date to show = result = 02/09/2017.
But I still think how to do it if you can guide me I would appreciate a world, and everything should show without pressing buttons, if I press a button would be like to store in the database only the results otenidos as date1 date2 and result
<? Php
// $ date1 = $ _ POST ["date1"]; // this will be the first date or date of entry
// $ date2 = $ _ POST ["date2"]; // this will be the date with which you will calculate
// $ difference = $ date2 - $ date1;
// if ($ difference <= 5)
{
// echo "number of days that correspond to it [". $ Date2 + 30 + 3. "].";
//} else {
// ($ difference <= 5)
{
// echo "number of days corresponding to it [" $ date2 + 30 "].";
}
stoicnluv at gmail dot com ¶
14 years ago
A better and accurate function to calculate the difference between 2 dates. Takes leap years and DST into consideration. Accepts string date or timestamp as arguments.
<?php
function date_diff($d1, $d2){
$d1 = (is_string($d1) ? strtotime($d1) : $d1);
$d2 = (is_string($d2) ? strtotime($d2) : $d2);
$diff_secs = abs($d1 - $d2);
$base_year = min(date("Y", $d1), date("Y", $d2));
$diff = mktime(0, 0, $diff_secs, 1, 1, $base_year);
return array(
"years" => date("Y", $diff) - $base_year,
"months_total" => (date("Y", $diff) - $base_year) * 12 + date("n", $diff) - 1,
"months" => date("n", $diff) - 1,
"days_total" => floor($diff_secs / (3600 * 24)),
"days" => date("j", $diff) - 1,
"hours_total" => floor($diff_secs / 3600),
"hours" => date("G", $diff),
"minutes_total" => floor($diff_secs / 60),
"minutes" => (int) date("i", $diff),
"seconds_total" => $diff_secs,
"seconds" => (int) date("s", $diff)
);
}
$a = date_diff("2006-11-01", "2007-11-01");
echo "<pre>";
print_r($a);
echo "</pre>";
?>
This example will output (if your timezone uses US DST):
Array
(
[years] => 0
[months_total] => 11
[months] => 11
[days_total] => 364
[days] => 30
[hours_total] => 8759
[hours] => 23
[minutes_total] => 525540
[minutes] => 0
[seconds_total] => 31532400
[seconds] => 0
)
As you can see, the result is not exactly 1 year (less 1 hour) since Nov 1, 2006 is not DST while Nov 1, 2007 is DST.
aquatakat at telus dot net ¶
14 years ago
I wrote a simple script to format a duration in seconds. Give the function some value in seconds and it will return an array.
<?php
function format_duration($seconds) {
$periods = array(
'centuries' => 3155692600,
'decades' => 315569260,
'years' => 31556926,
'months' => 2629743,
'weeks' => 604800,
'days' => 86400,
'hours' => 3600,
'minutes' => 60,
'seconds' => 1
);
$durations = array();
foreach ($periods as $period => $seconds_in_period) {
if ($seconds >= $seconds_in_period) {
$durations[$period] = floor($seconds / $seconds_in_period);
$seconds -= $durations[$period] * $seconds_in_period;
}
}
return $durations;
}
echo format_duration(864);
/*
[minutes] => 14
[seconds] => 24
*/
echo format_duration(3600);
/*
[hours] => 1
*/
echo format_duration(11111111);
/*
[months] => 4
[days] => 6
[hours] => 20
[minutes] => 28
[seconds] => 59
*/
?>
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