Re: Variable Undefined, even when global?

You are a god among men, Ernest. THANK YOU! -ron "Ernest E Vogelsinger" <ernest@vogelsinger.at> wrote in message news:5.1.1.6.2.20030309014055.031e93a8@mail.vogelsinger.at... > At 01:28 09.03.2003, Ron Biggs said: > --------------------[snip]-------------------- > >function ColorCollection($lkID){ > > > > $whatX = $lkID + 1; // Adding 1 turns makes the index correspond to the > >template number. > > > >// $aColLen = count($lkColor); //.length > > $aColLen = test; //sizeof($lkColor); //.length > > > > echo "<option value=\"\" selected>xxx</option>\n"; //first > > option is blank > > > >// KeepGoing: //PHP does not have a line-label statement. Set up a trigger > > for($i=0;$i<$aColLen;$i++){ > > $trigger = false; > > $whatglass = $lkColor[$i]->glass; //shouldn't $lkColor[any-number] be > >global??? <<<<<<<<<<< > --------------------[snip]-------------------- > > Yes it _is_ global, but within the function you need to declare if global > to have access to it, if you miss this you're referencing a local variable, > which is not defined... > > function ColorCollection($lkID){ > > global $lkColor; // <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< > > $whatX = $lkID + 1; // Adding 1 turns makes the index correspond to the > template number. > // $aColLen = count($lkColor); //.length > $aColLen = test; //sizeof($lkColor); //.length > echo "<option value=\"\" selected>xxx</option>\n"; //first > option is blank > // KeepGoing: //PHP does not have a line-label statement. Set up a trigger > for($i=0;$i<$aColLen;$i++){ > $trigger = false; > $whatglass = $lkColor[$i]->glass; //shouldn't $lkColor[any-number] be > global??? > > > > -- > >O Ernest E. Vogelsinger > (\) ICQ #13394035 > ^ http://www.vogelsinger.at/ > >