Re: using $$foo_name vs. $foo_name for file upload

Tony wrote:

Trying to use the following statement to upload a file via a form, when I use a variable "$foo" in the following statement it works correctly: if (copy($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?

Variable variables are a cheap man's array. Why not just name all of your file inputs the same and tack a "[]" onto the end of the name. Then you can treat $_FILES as a multi-dimensional array and just loop through it and process it. -- ---John Holmes... Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/ php|architect: The Magazine for PHP Professionals – www.phparch.com