Re: $VAR = 0 != false

Description int empty(mixed var); Returns false if var is set and has a non-empty or non-zero value; true otherwise. That means that if it's zero i will return true. You're using not (!) so it returns false. Stick with isset(), which is suitable for what you need. -- André Fadel fadel@fee.unicamp.br Hoje E Sempre GUARANI! On Mon, 7 Aug 2000, Brian T. Allen wrote: > Hi *, > > I have an ecommerce site where the shopping cart supports options for items. > > One of the items has an option that is stored in the database as 0 (zero). It doesn't > mean false, it means that an option was selected, and the option was 0. > > If I echo out the variable, you get "0" echoed out. > > $VAR = "0"; > echo $VAR; > > You get what you would expect. > > But, if you verify the variable exists first: > > if($VAR){ > echo $VAR; > } > > You get nothing, even though the variable is set. It is set to zero, granted, but it is set! > > I can understand that in some types of tests you want 0 to be synonimous with false, but here I > am only testing if the variable is set to a value, be it false, 0, or something else. > > I just tested, and isset() works like I think it should (I am NOT the source of all truth and > knowledge, just trying to make this make sense), but !empty() does not: > > if(!empty($VAR)){ > echo $VAR; > } > > even though $VAR isn't empty. > > Thoughts, comments, flames? > > Thanks, > Brian > >