print $text as text ($text is not a variable)


I want to do this


but it just try’s to print the variable $text which doesn’t exist


Well, just “escape” the question mark, like this: print("$text");
Or, even print("$" . “text”); (so the variable does not exist.)


Thanks, sorry for my silly questions, I’m completely new programming in all languages, only started this week… ill remember this one.



LOL, no question is silly! Just sometimes it is right in front of your face… Always nice to hear “SOLVED!”

CYA in the bitstream…