Question

Given the following file/directory structure:

/photos/1/original/filename1.jpg /photos/1/thumb/filename1.jpg /photos/2/original/filename2.jpg /photos/2/thumb/filename2.jpg /photos/3/original/filename3.jpg /photos/3/thumb/filename3.jpg 

…etc. I’m looking for the regexp to convert the file/directory structure to:

/photos/1/original.jpg /photos/1/thumb.jpg /photos/2/original.jpg /photos/2/thumb.jpg /photos/3/original.jpg /photos/3/thumb.jpg 

…etc. I’ve got scripts to do the file renaming in ruby or perl…just missing the proper regexp.

Answer 1

Perl solution

$newfn = $oldfn; $newfn =~ s/(original|thumb)[^\.]*/$1/; rename($oldfn, $newfn); 

As requested, to match any keyword, not just ‘original’ or ‘thumb’, assuming that the /photos/nn remains the same:

$newfn = $oldfn; $newfn =~ s?(photos/\d+/[^/]*)/[^\.]*?$1?; rename($oldfn, $newfn); 

Adapting this solution to also remove the directories if they are empty

$newfn = $oldfn; $newfn =~ s?(/photos/\d+/)([^/]*)/[^\.]*?$1$2?; rename($oldfn, $newfn); unlink($1$2); 

I’m not sure what this will do if the directory is not empty when you go to unlink it.