if (copy($foo, "./$foo_name")) {//blah... }The name of the file I am trying to upload $foo is evaluated correctly. Likewise I can obtain the file size, "$foo_size".
However, it would be advantageous (for reasons that are to complicated to explain here) to use a "variable of a variable" such as "$$foo" however this does not work...
if (copy($$foo, "./$$foo_name")) {//blah... }From my testing it appears that the problem has to do with the evaluation of "./$$foo_name". I have tried using a substitute variable such as
$substitute = $$foo;
if (copy ($substitute, "./$substitute_name")) {//blah... }but that does not work either.
Can someone explain how I might get the file name from a $$variable?
Tony
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