On 11/2/05, Unknown Unknown <phpinfolist@xxxxxxxxx> wrote: > > Hi everybody i have a class that i reference with: > $DB= new DBInterface; > outside a function it works fine, but using $DB inside a function changes > the data type i think... i get an error saying i'm using a method on a > non-object... > any help appreciatted if you mean by "inside a function", a method of your class DBInterface then you can can use $this. Otherwise you can do something like: $DB= new DBInterface(); and then I believe you should be able to use $DB at it is.
