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wednesday wrote

fwiw, that's not a correct use of escapeshellcmd. it doesn't modify the string in-place, it returns the escaped string, so you would need something like $safe_filename = escapeshellcmd($filename).

however in this case i think you actually want escapeshellarg instead, since the filename is a single argument.


Delonix wrote

What to do unordered to understand code like u guys