John W. Holmes wrote:
What John has suggested is the most common way of doing it, if you insist on using variable variables you might want to take a look at evalTony 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.
you might also want to echo out $substitute to see what it contains.
-- Raditha Dissanayake. --------------------------------------------- http://www.raditha.com/megaupload/upload.php Sneak past the PHP file upload limits.
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