PHP 5.4.36 Released

Date/Time 函数

Table of Contents

add a note add a note

User Contributed Notes 24 notes

up
7
koch.ro
7 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
       
);
    }
}
?>
up
3
th at definitynet dot com
13 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!
up
4
andreencinas at yahoo dot com dot br
9 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
        
}

       }
?>
up
1
bgold at matrix-consultants dot com
7 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";
                }
        }
?>
up
1
php-contrib at i-ps dot nospam dot net
12 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.
up
1
daniel at globalnetstudios dot com
9 years ago
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;
  }
?>
up
1
mail at completeideas dot com
9 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])
up
1
mincklerstraat at softhome dot net
10 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 .
up
1
php at sarge dot ch
10 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';
      }
  }
}
?>
up
1
php at elmegil dot net
11 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.
up
1
garyc at earthling dot net
11 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
?>
up
1
nightowl at NOS-PA-M dot uk2 dot net
12 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;
   
}
?>
up
0
Robb_Bean at gmx dot nospam dot net
6 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
up
0
venoel at rin dot ru
7 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.
up
0
Leopoldo A dot Oducado (poducado at comfsm dot fm)
7 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");
?>
up
0
glashio at xs4all dot nl
9 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;
}
?>
up
0
brighn (a) yahoo (.) com
11 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";
  }
?>
up
-1
Hans
3 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!
}

?>
up
-1
Darren Edwards
7 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.
up
-1
nickaubert at america's biggest isp dot com
10 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...
 
}
?>
up
-2
luck dot lil dot leprechaun at gmail dot com
5 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.]
up
-2
sagar
6 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
?>
up
-3
stoicnluv at gmail dot com
7 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.
up
-3
aquatakat at telus dot net
7 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
*/

?>
To Top