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Calculate age from two dates

Thread began 6/26/2009 12:11 pm by Daryl | Last modified 6/26/2009 1:57 pm by Justin Nemeth | 4172 views | 2 replies |

DarylBeta Tester

Calculate age from two dates

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.

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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(000$mon1$day1$year2-1))/(24*60*60));                        
                    } else {
                        
$yddiff intval(($end mktime(000$mon1$day1$year2))/(24*60*60));
                    }
                    print(
$yddiff);
                    break;
                case 
"MD":
                    if(
$day1>$day2) {
                        
$mddiff intval(($end mktime(000$mon2-1$day1$year2))/(24*60*60));                        
                    } else {
                        
$mddiff intval(($end mktime(000$mon2$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"); 
?>
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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.

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