$days = 35; $day = $day + $days; //calculate the new date $calcDate = date($output, mktime (0,0,0,$month,$day,$year)); No need for a 2nd query...jus' insert the variable $calcDate into a column in your table Hope this helps, Jimmy Brock "Shannon Doyle" <shannon@xxxxxxxxxxxxxxxxx> wrote in message news:200403211314.i2LDE83B030712@xxxxxxxxxxxxxxxxxxxxxxxx > Hi People, > > Need some assistance in the following scenario:- > > Inserting a date into the database that is entered into a form by the site > visitor. - This is easy enough. > > However I now need to use the same date that has been entered by the site > visitor add 35 days and then insert into another table. > > My question, how do I get the date entered into the form add 35days to it > and then include that into the same sql query as the first one. Or do I have > to use a second sql query? If the second query how would I get the date and > add 35days?? > > > Cheers, > > Shannon -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
