I found this code at Google's cached version of the manual for the date() function: http://216.239.37.104/search?q=cache:kq0aNfZeEp8J:www.php.net/date+diffe rence+two+date+site:www.php.net&hl=en&ie=UTF-8 This code only returned the number of hours, minutes, and seconds between the two points in time. Now it returns years and days as well as the left over time. The script also used the mod operator (%) to find the remaining days after determining the year and it would not calculate correctly. For example, if I compared yesterday, three years ago(08-01-00) and today (08-02-03) at noon for both days, I would get a difference of 3 years. No I am using while loops subtracting whole years to find the number of days that remain. The same is true for subtracting days, hours and minutes. CODE: <?php $diff = dateDiff("12-21-1983 10:20:00", "08-02-2003 16:42:01"); echo "$diff\r\n"; function unixTm($strDT) { $arrDT = explode(" ", $strDT); $arrD = explode("-", $arrDT[0]); $arrT = explode(":", $arrDT[1]); return mktime($arrT[0], $arrT[1], $arrT[2], $arrD[0], $arrD[1], $arrD[2]); } function dateDiff($date1,$date2) // dateDiff(date_in_the_past, date_in_the_future) { $dt2=unixTm($date2) ; $dt1=unixTm($date1) ; $r = $dt2 - $dt1; $yy=floor($r / 31536000); while ($r > 31536000){ $r -= 31536000; } $dd=floor($r / 86400); while ($r > 86400){ $r -= 86400; } $hh=floor($r/3600); if ($hh<=9) $hh="0".$hh; //adds a leading 0 if less than 10 while ($r > 3600){ $r -= 3600; } $mm=floor($r/60); if ($mm<=9) $mm="0".$mm; //adds a leading 0 if less than 10 while ($r > 60){ $r -= 60; } $ss=$r ; if ($ss<=9) $ss="0".$ss; //adds a leading 0 if less than 10 $retval="$yy year(s), $dd day(s) $hh:$mm:$ss"; return $retval; } ?> OUTPUT: 19 year(s), 229 day(s) 05:22:01 Hope this helps, Ryan Marks -----Original Message----- From: John Ryan [mailto:celticfc@iol.ie] Sent: Saturday, August 02, 2003 2:31 PM To: php-general@lists.php.net; php-db@lists.php.net Subject: subtracting dates... Hi, In mySQL, I store dates as YYYY-MM-DD, a standard DATE type. It stores users date of births. I need to calculate in a PHP script, the users age from this DOB. I get a PHP date in the same format as the mySQL and subtract, which returns the year rounded off. ie, it doesnt matter if your birthdays in june of 1983 and the date is januray 2003, your age is still returned as 20, when it should be 19. Does anyone know how can i get the right age? -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
