2005/8/24, Cummings, Shawn (GNAPs) <[EMAIL PROTECTED]>: > > The 29th wouldn't be an issue because if that is their birthday -- and > today is 2/29 -- it will show up.
It will only happen once every 4 years... I'll go with the two fields solution and make a special case for leap years. thanks for your help. > > > Pooly wrote: > > >Hi, > > > >I would like to display a list of members who have their birthday a > >given day (today for instance). > >My idea is to store their birth date in a column, and then query the > >table against the column. But the query would be like : > >select id from members where MONTH(birthday) = MONTH(NOW()) AND > >DAY(birthday)=DAY(NOW()) > >but it would perform a entire table scan with that. > >What would be your best strategy for that sort of query ? > >And how would you deal with 29th of february ? > > > > > > > > > -- Pooly Webzine Rock : http://www.w-fenec.org/ -- MySQL General Mailing List For list archives: http://lists.mysql.com/mysql To unsubscribe: http://lists.mysql.com/[EMAIL PROTECTED]
