Re: Variable in $_FILES?

Just to answer my own question, but for reference for anyone else who has this problem, I made a simple error (not very experienced with PHP). Anyway, turns out that 'object$objectNumber' is different than 'object'$objectNumber and also different that just $objectNumber. As most of you know, putting ' ' around a variable makes PHP print it out exactly like that (ie. '$objectNumber' = $objectNumber in PHP) whereas I just wanted the variable properties to actually be put into the $_FILES function (or whatever it is). I couldn't figure out how to put 'object'$objectNumber together so I just changed it to simply $objectNumber, so it now works. This is what the code is now: $file = $_FILES{$objectNumber}; $fileName = $file['name']; $filetmp = $file['tmp_name']; ... and everything works. Dimitri Marshall "Dimitri Marshall" <webmaster@dyntdesign.com> wrote in message news:20031202004839.86219.qmail@pb1.pair.com... > Hi there, > This is what I'm trying to do: > > $pic = $_FILES['object$objectNumber']; > $picName = $pic['name']; > $pictmp = $pic['tmp_name']; > > $objectNumber is being defined, but for some reason this code won't work. > Can someone tell me why and also what I can do to make it work? > > Thanks in advance, > Dimitri Marshall