I wanted to simply print out these details for a directory, and I wanted to write the script in perl, any insights?
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Use stat
to get the UID of the file's owner and getpwuid
to get the username for the ID, e.g.:
my $owner = getpwuid((stat)[4]);
Note that, if you call getpwuid
in list context, it will return a list of values, the first of which is the username.

jwodder
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Is there any way , I could do something similar, but access file version? – salem May 17 '11 at 23:23
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If by "file version" you mean the version of the application used to create/open/modify the file, that information is stored within the file itself, and extracting it depends on the specific type of file. – jwodder May 17 '11 at 23:25
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I'm using this script only one applications, not on actual files, so I need to get the version of say an executable, how would I do that? – salem May 17 '11 at 23:28
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It depends entirely on the executable. By convention, running the executable with the `-v`, `-V`, or `--version` switch will produce version information, but not all programs support this, and the format of the output varies between programs. – jwodder May 17 '11 at 23:30
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1I think salem is talking about the standard Windows metadata embedded into executable files (see properties dialog in Explorer). – daxim May 18 '11 at 11:19