I need to build a list of files present in a directory but also pass in a file containing a list of directories to exclude.

I've seen the below suggestion but I've had no luck with it when it comes to the type flag:

find $dir_name $(printf "! -name %s " $(cat $exclude_file))

Is there a way I can do it or will I need to create the entire list and then work through it removing ones that partially match ones in the exclude directory.

  • 4
    What do you mean by "I've had no luck with it when it comes to the type flag"? Can you show the exact command you're running that doesn't work? – Michael Mol Apr 10 '17 at 14:26

Given that your exclude_file contains paths, not names, you need to use -path to match its entries. To exclude matching directories’ subdirectories, you also need to -prune them. This should work:

find . -type d \( $(printf -- "-path */%s -o " $(cat "$exclude_file")) -false \) -prune -o -print

If you only want to see files, you can print only files:

find . -type d \( $(printf -- "-path */%s -o " $(cat "$exclude_file")) -false \) -prune -o -type f -print
  • I've edited this answer, you needed -- in-between printf "-path", or you'll get an "Invalid option -p" due to printf trying to parse it. – Someguy123 Oct 9 '18 at 11:17
  • Thanks @Someguy123; I tested with zsh, which doesn’t need --, but bash does indeed. – Stephen Kitt Oct 9 '18 at 11:22

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