Re: more an SQL issue but I can't find any help

On Fri, 20 Apr 2001 17:02, you wrote: > Yes but how I can "say" 18 years ??? > > Marian Vasile > IT Manager > Schnecker van Wyk & Pearson > www.investments.ro > +40 (0) 1 2309000 > > > -----Original Message----- > > From: David Robley [mailto:huntsman@www.nisu.flinders.edu.au] > > Sent: Friday, April 20, 2001 5:31 AM > > To: Marian Vasile; php-general@lists.php.net > > Subject: Re: [PHP] more an SQL issue but I can't find any help > > > > On Fri, 20 Apr 2001 11:35, Marian Vasile wrote: > > > I have a table with users and their birthdates. > > > I want to SELECT all the users who have more than 18 years. How I > > > can do that using bisect years ? (february 28 days and 29 days) > > > > > > plz help ... > > > > > > Marian Vasile > > > IT Manager > > > Schnecker van Wyk & Pearson > > > www.investments.ro > > > +40 (0) 1 2309000 > > > > Leap years shouldn't be a problem? I don't know what DB you are using > > or what format you store your dates - but if you can do a query > > something like > > > > SELECT * FROM table WHERE birthdate > (NOW()-18years) > > > > using whatever your DB supports for the last bit :-) I _think_ it > > might work as is with recent MySQL. Well, it depends on your DB - for Mysql look for the date_sub or subdate functions; for anything else, you'll have to RTM :-0 -- David Robley | WEBMASTER & Mail List Admin RESEARCH CENTRE FOR INJURY STUDIES | http://www.nisu.flinders.edu.au/ AusEinet | http://auseinet.flinders.edu.au/ Flinders University, ADELAIDE, SOUTH AUSTRALIA