Calculate age from two dates

6/26/2009 9:11 am | #1 DarylBeta Tester

I would like to calculate an age of a person between two different date. The dates are both in yyyy-mm-dd

I am using the following;

$row_member_profile['dog_dob']
$row_member_profile['years']

I have tried lots of different things but just can't get it to work.

Any help would be greatly appreciated.

6/26/2009 10:37 am | #2 DarylBeta Tester

I managed to solve it myself so thought I would post my solution for anyone else who might like to do this;

I created a file called datediff.php which contained the following;

php:

<?php
    function datediff($start_date,$end_date,$unit="y")
        {
            $unit = strtoupper($unit);
            $start=strtotime($start_date);
            if ($start === -1) {
                print("invalid start date");
            }
            
            $end=strtotime($end_date);            
            if ($end === -1) {
                print("invalid end date");
            }
            
            if ($start > $end) {
                $temp = $start;
                $start = $end;
                $end = $temp;
            }
            
            $diff = $end-$start;
            
            $day1 = date("j", $start);
            $mon1 = date("n", $start);
            $year1 = date("Y", $start);
            $day2 = date("j", $end);
            $mon2 = date("n", $end);
            $year2 = date("Y", $end);
            
            switch($unit) {
                case "D":
                    print(intval($diff/(24*60*60)));
                    break;
                case "M":
                    if($day1>$day2) {
                        $mdiff = (($year2-$year1)*12)+($mon2-$mon1-1);
                    } else {
                        $mdiff = (($year2-$year1)*12)+($mon2-$mon1);
                    }
                    print($mdiff);
                    break;
                case "Y":
                    if(($mon1>$mon2) || (($mon1==$mon2) && ($day1>$day2))){
                        $ydiff = $year2-$year1-1;
                    } else {
                        $ydiff = $year2-$year1;
                    }
                    print($ydiff);
                    break;
                case "YM":
                    if($day1>$day2) {
                        if($mon1>=$mon2) {
                            $ymdiff = 12+($mon2-$mon1-1);
                        } else {
                            $ymdiff = $mon2-$mon1-1;
                        }
                    } else {
                        if($mon1>$mon2) {
                            $ymdiff = 12+($mon2-$mon1);
                        } else {
                            $ymdiff = $mon2-$mon1;
                        }
                    }
                    print($ymdiff);
                    break;
                case "YD":
                    if(($mon1>$mon2) || (($mon1==$mon2) &&($day1>$day2))) {
                        $yddiff = intval(($end - mktime(0, 0, 0, $mon1, $day1, $year2-1))/(24*60*60));                        
                    } else {
                        $yddiff = intval(($end - mktime(0, 0, 0, $mon1, $day1, $year2))/(24*60*60));
                    }
                    print($yddiff);
                    break;
                case "MD":
                    if($day1>$day2) {
                        $mddiff = intval(($end - mktime(0, 0, 0, $mon2-1, $day1, $year2))/(24*60*60));                        
                    } else {
                        $mddiff = intval(($end - mktime(0, 0, 0, $mon2, $day1, $year2))/(24*60*60));
                    }
                    print($mddiff);
                    break;
                default:
                 print("{Datedif Error: Unrecognized \$unit parameter. Valid values are 'Y', 'M', 'D', 'YM'. Default is 'D'.}");
                
            }

        }

?>

Then inlcuded it;

php:

<?php 
include('dateDiff.php'); 
?>

Then added the code to my page;

php:

<?php 
          $date_1 = $row_member_profile['dog_dob'];
          $date_2 = $row_member_profile['years'];
          datediff ("$date_1","$date_2","y"); 
?>

6/26/2009 10:57 am | #3 Justin Nemeth

If you are using MySQL (which I assume you are based on the $row variable), check out the datediff function.

date-and-time-functions.html#function_datediff

That will give you the number of days between 2 dates. So you could do a statement like this to get the age.

SELECT (DATEDIFF(dog_dob, years) / 365) AS age ....

As for PHP, looks like your function will do what is needed, but it is probably overkill if you only want the years. Something like this would work. Basically determine the number of seconds between the dates, then divide by the number of seconds in a year.

php:

<?php 
$date_1 = $row_member_profile['dog_dob'];
$date_2 = $row_member_profile['years'];

$seconds = abs(strotime($date_1) - strotime($date_2));
$years = floor($seconds / (365 * 24 * 60 * 60));

print $years;
?>

I woud say the MySQL route is the easiest to do.

Technical Support Forums

Free, outstanding support from WebAssist and your colleagues

Calculate age from two dates

6/26/2009 9:11 am | #1 DarylBeta Tester