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@NogDogAug 30.2006 — #
Try that with at birthdate of 1986-12-31, and you'll find that it gives the same age, even though the person hasn't had his/her birthday yet for this year. That is what the more complicated function you were looking at is trying to deal with.
(The above function is actually just subtracting the year portions of each value, as anything in the strings from the first non-numeric character (the hyphen) will be ignored when treating strings as numbers.)@NogDogAug 30.2006 — #The gregoriantojd() function wants its arguments in this order: month, day, year. So if your dates are in y-m-d (which is much better, in my opinion), then you'll need to change these two lines:
php myql age
HI
I’m trying to calculate a persons age. Their date of birth is stored in a mysql table (column is set as date) format yyyy-m-d
how would I calculate their age based on todays date
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@NogDogAug 30.2006 — #[code=php]<?
$currentDate = date('Y-m-d');
$birthDate = "1986-07-22";
$age = $currentDate - $birthDate;
echo $age;
?>[/code]
for the var $birthDate you would set that to what is in the database, or where ever your getting the source from.[/QUOTE]
Try that with at birthdate of 1986-12-31, and you'll find that it gives the same age, even though the person hasn't had his/her birthday yet for this year. That is what the more complicated function you were looking at is trying to deal with.
(The above function is actually just subtracting the year portions of each value, as anything in the strings from the first non-numeric character (the hyphen) will be ignored when treating strings as numbers.)
0
@NogDogAug 30.2006 — #The gregoriantojd() function wants its arguments in this order: month, day, year. So if your dates are in y-m-d (which is much better, in my opinion), then you'll need to change these two lines:[code=php]
$start_date=gregoriantojd($date_parts1[1], $date_parts1[2], $date_parts1[0]);
$end_date=gregoriantojd($date_parts2[1], $date_parts2[2], $date_parts2[0]);
[/code]