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I'm looking for a shell command in order to get all the files in a complex directory structure. By complex directory structure I mean that there's a root folder with more than 150 subfolders and for each subfolder there are minimum 3 subfolders.

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What do you want to do with the list of files? What you want to do with the list affects the best way to get it. –  depquid Mar 15 '13 at 18:44
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migrated from serverfault.com Mar 15 '13 at 16:45

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4 Answers

This is trivial:

find /directory -type f
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As suggested by @michael-hampton find is the way to go. However a bit more explanation is needed. The option -type can have several values, check the man page (man find). Here is a quick definition:

If you are looking for regular files only:

find <path> -type f

If you are looking for anything not a directory:

find <path> ! -type d

If you are looking for regular files and symbolic links:

find <path> -type f -o -type l

(the previous command looks for regular OR link)

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find works just as well. But if you are using zshell (zsh):

For only files:

ls -al path/**/*(.)

For only dirs:

ls -al path/**/*(/)

See man zshexpn for more eamples. Specifically the "glob qualifier" section.

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You can also do this in bash4+ (be aware that this follows symlinks):

shopt -s globstar nullglob
for file in **/*; do
    [[ -f $file ]] && printf '%s\n' "$file"
done
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The silly code coloring makes me want to put a comment # */ just to fix it. –  kojiro Mar 15 '13 at 22:37
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@kojiro - Sorry, forgot to force the syntax highlighter to bash mode. –  Chris Down Mar 16 '13 at 23:53
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Now I'm exceedingly happy I posted that dumb comment, because I didn't know you could force the highlighter to bash mode. –  kojiro Mar 17 '13 at 0:34
1  
@kojiro, in case you not found it yet, see Syntax highlighting for code in the Markdown help. –  manatwork Mar 19 '13 at 7:27
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