Posted by walterbyrd on 12/17/88 11:54
I know I can use the file() function to read a file into an array by providing a literal file name between single quotes: $lines = file('literal_file_name'); But, what if I want file() to use a variable name? $variable_file_name = "literal_file_name"; $lines = file($variable_file_name); Of course, that example won't work. But what will?
[Back to original message]
Copyright © 2005-2006 Powered by Custom PHP Programming