* [EMAIL PROTECTED] <[EMAIL PROTECTED]> : > I am not sure > but maybe something like this is posible > > $current_year = date(Y); > $qry = "SELECT * FROM > `table` > WHERE ($current_year-YEAR(birthday)) BETWEEN 25 AND 26"
No -- because this incorrectly identifies somebody born on June 14 as 26 even though today is May 9, and they're still 25. > "Matthew Weier O'Phinney" <[EMAIL PROTECTED]> wrote in message > news:[EMAIL PROTECTED] > > * Ryan A <[EMAIL PROTECTED]> : > > > Thanks for replying. > > > > > > > SELECT * FROM > > > > `table` > > > > WHERE `age` BETWEEN 25 AND 26; > > > > > > I knew the above, but how do i use it with my date field when i have > > > birthdates like this: > > > 01-01-1969 > > > and 03-05-1955 -- Matthew Weier O'Phinney | WEBSITES: Webmaster and IT Specialist | http://www.garden.org National Gardening Association | http://www.kidsgardening.com 802-863-5251 x156 | http://nationalgardenmonth.org mailto:[EMAIL PROTECTED] | http://vermontbotanical.org -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
