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    3
    daniel at globalnetstudios dot com
    19 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;
      }
    ?>
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    3
    mail at completeideas dot com
    19 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])
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    3
    mincklerstraat at softhome dot net
    20 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
    3
    bgold at matrix-consultants dot com
    17 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";
                    }
            }
    ?>
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    2
    nickaubert at america's biggest isp dot com
    20 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...
          }
    ?>
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    1
    brighn (a) yahoo (.) com
    21 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";
      }
    ?>
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    1
    php-contrib at i-ps dot nospam dot net
    22 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.
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    0
    nightowl at NOS-PA-M dot uk2 dot net
    22 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;
       
    }
    ?>
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    -1
    garyc at earthling dot net
    21 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
    ?>
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    -1
    th at definitynet dot com
    23 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
    -2
    koch.ro
    17 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
                );
        }
    }
    ?>
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    -2
    Hans
    13 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!
    }

    ?>
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    -3
    andreencinas at yahoo dot com dot br
    19 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
                 }

           }
    ?>
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    -3
    venoel at rin dot ru
    17 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
    -2
    Darren Edwards
    17 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.
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    -3
    luck dot lil dot leprechaun at gmail dot com
    15 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.]
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    -3
    php at elmegil dot net
    21 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.
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    -4
    sagar
    16 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
    -4
    Leopoldo A dot Oducado (poducado at comfsm dot fm)
    17 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
    -5
    glashio at xs4all dot nl
    19 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
    -4
    Robb_Bean at gmx dot nospam dot net
    16 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
    -3
    php at sarge dot ch
    20 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
    -5
    paez903 at hotmail dot com
    7 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 "].";

    }
    up
    -5
    stoicnluv at gmail dot com
    17 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
    -7
    aquatakat at telus dot net
    17 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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