Scripting: Skipping n files

chef_carol

I'm trying to write a rather simple script, but can't find a reasonable approach to skip some files. I'd like to list all files somewhere with a given order. if that listing contains more then (lets say) 10 files, I want to remove files 11-n but leave the first (or last) 10 in place. /TLDR: How do I skip a fixed amount of files in a find/ls/whatever result before piping into the next command I want to sort all files (with given pattern) by name...

Originally asked by Jaster on Stack Exchange (unix) (CC BY-SA 4.0).

uphilltech

Great question — answering as the UphillTech team:

Using zsh , the pattern *(.[11,-1]) would match all names of regular files (not directory names etc.) in the current directory, sorting them by name (the default sort order), and would return all but the first ten of these names. Testing: $ touch file-{01..20} $ ls file-01 file-03 file-05 file-07 file-09 file-11 file-13 file-15 file-17 file-19 file-02 file-04 file-06 file-08 file-10 file-12 file-14 file-16 file-18 file-20 $ printf '%s\n' *(.[11,-1]) file-11 file-12 file-13 file-14 file-15 file-16 file-17 file-18 file-19 file-20 Would you want to delete these files: $ rm *(.[11,-1]) $ ls file-01 file-02 file-03 file-04 file-05 file-06 file-07 file-08 file-09 file-10 Note that this requires...

Full answer by Kusalananda: https://unix.stackexchange.com/a/515907 (CC BY-SA 4.0).

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