I might be wrong - but isn't that going to mess up on or around the birthday
because of leap years ?
I know its more complex - but something like :
select year(curdate())-year("1952-06-24")-
(dayofyear(curdate())<dayofyear("1952-06-24")) age;
might work better...
On Monday 25 June 2007 21:07, Mike Blezien wrote:
> thanks, worked prefectly,
>
> Mike
> ----- Original Message -----
> From: "Jerry Schwartz" <[EMAIL PROTECTED]>
> To: "'Mike Blezien'" <[EMAIL PROTECTED]>; "'MySQL List'"
> <mysql@lists.mysql.com>
> Sent: Monday, June 25, 2007 3:01 PM
> Subject: RE: Birthday format
>
> > mysql> SELECT FLOOR(54.75);
> > +--------------+
> >
> > | FLOOR(54.75) |
> >
> > +--------------+
> >
> > | 54 |
> >
> > +--------------+
> > 1 row in set (0.05 sec)
> >
> > Regards,
> >
> > Jerry Schwartz
> > The Infoshop by Global Information Incorporated
> > 195 Farmington Ave.
> > Farmington, CT 06032
> >
> > 860.674.8796 / FAX: 860.674.8341
> >
> > www.the-infoshop.com
> > www.giiexpress.com
> > www.etudes-marche.com
> >
> >> -----Original Message-----
> >> From: Mike Blezien [mailto:[EMAIL PROTECTED]
> >> Sent: Monday, June 25, 2007 2:48 PM
> >> To: MySQL List
> >> Subject: Birthday format
> >>
> >> Hello,
> >>
> >> we have a simple query to calculate someones birthday:
> >>
> >> SELECT (TO_DAYS(CURDATE()) - TO_DAYS('1952-10-06')) / 365;
> >>
> >> = 54.75
> >>
> >>
> >> Now is there away, using SQL, to remove the .75 without
> >> rounding off. we just
> >> want the '54' value. We can do it easy enough with our Perl
> >> programming, but was
> >> wondering if this can be accomplished within the actual query itself.
> >>
> >> TIA,
> >>
> >> Mike(mickalo)Blezien
> >> ===============================
> >> Thunder Rain Internet Publishing
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> >> http://www.thunder-rain.com
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> >>
> >> --
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