Converting Date to Age

andretanguy

Hello there,

I'm trying to convert a users date in the database into their age using the following:

<?php

//calculate years of age (input string: YYYY-MM-DD)
function dateDiff($dformat, $endDate, $beginDate)
{
$date_parts1=explode($dformat, $beginDate);
$date_parts2=explode($dformat, $endDate);
$start_date=gregoriantojd($date_parts1[0], $date_parts1[1], $date_parts1[2]);
$end_date=gregoriantojd($date_parts2[0], $date_parts2[1], $date_parts2[2]);
return $end_date - $start_date;
}

$date1 = date('d-m-Y');
$date2 = $row_filteredUsersInRange['date_birthday'];

print " . dateDiff("-", $date2, $date1) . ";

However, the age always appears as 0 and then i get the error: Fatal error: Cannot redeclare datediff() (previously declared in /home/hacs/FRSI2ZHA/htdocs/postcode-uk/postcode-results.php:285) in /home/hacs/FRSI2ZHA/htdocs/postcode-uk/postcode-results.php on line 285

Line 285: function dateDiff($dformat, $endDate, $beginDate)

Also, the date in the database is in the format: YYYY-MM-DD 00:00:00

If anyone can help that would be great

rocklv

function dateDiff2($dformat, $endDate, $beginDate)
{
$date_parts1=explode($dformat, $beginDate);
$date_parts2=explode($dformat, $endDate);
$start_date=gregoriantojd($date_parts1[0], $date_parts1[1], $date_parts1[2]);
$end_date=gregoriantojd($date_parts2[0], $date_parts2[1], $date_parts2[2]);
return abs($end_date - $start_date)/365;
}

$date1 = date('n-j-Y');
$date2 = "7-22-1983";

print dateDiff2("-", $date2, $date1); 
print " years old";

Roger_Ramjet

You can do this all in the query using the mysql date functions

date_diff returns the number of days between 2 dates, divide by 365 for years of age. Problem is that leap years will throw some age calculations out.

period_diff will give you the number of months between two dates and gets around the leap year error. The problem here is if it is the month of their birthday you may get rounded up or down wrongly for their true age.

Working with unix timestamps incurs all the same problems with leap years. (and has the other problem of the seconds they add to the clock every few years to keep in sync with the sun - not relevent here but still a problem with unix timestamps)

I'd go with the period_diff myself


$sql = "SELECT period_diff(date_format(currdate(), '%Y%m'), date_format(dob, '%Y%m'))/12 AS age FROM users";