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and $src is the full path. What's the way to output only the filename (thus crop directories (ie crop the last '/' and everything before)) that take the least processing, or close to?
The script does more than only what's in the OP and I assume if I did basename instead of however the file names are currently fetch (full path) that it would break the script.
This chops from the left end.
It chops the leading common directory (start directory).
The following chops from the left end, and lets the * expand to the maximum (## not # where * expands to the minimum).
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