How to sort the results of find (including nested directories) alphabetically in bash

Question

I have a list of directories based on the results of running the \"find\" command in bash. As an example, the result of find are the files:

test/a/file
test/b/fi         


        

Answer 1

If you want to sort alphabetically, the best way is:

find test -print0 | sort -z

(The example in the original question actually wanted files before directories, which is not the same and requires extra steps)

Answer 2

try this. for reference, it firsts sorts on the second field second char. which only exists on the file, and has a r for reverse meaning it is first, after that it will sort on the first char of the second field. [-t is field deliminator, -k is key]

find test -name file |sort -t'/' -k2.2r -k2.1

do a info sort for more info. there is a ton of different ways to use the -t and -k together to get different results.

Answer 3

If you have the GNU version of find, try this:

find test -type f -printf '%h\0%d\0%p\n' | sort -t '\0' -n | awk -F '\0' '{print $3}'

To use these file names in a loop, do

find test -type f -printf '%h\0%d\0%p\n' | sort -t '\0' -n | awk -F '\0' '{print $3}' | while read file; do
    # use $file
done

The find command prints three things for each file: (1) its directory, (2) its depth in the directory tree, and (3) its full name. By including the depth in the output we can use sort -n to sort test/file above test/a/file. Finally we use awk to strip out the first two columns since they were only used for sorting.

Using \0 as a separator between the three fields allows us to handle file names with spaces and tabs in them (but not newlines, unfortunately).

$ find test -type f
test/b/file
test/a/file
test/file
test/z/file
$ find test -type f -printf '%h\0%d\0%p\n' | sort -t '\0' -n | awk -F'\0' '{print $3}'
test/file
test/a/file
test/b/file
test/z/file

If you are unable to modify the find command, then try this convoluted replacement:

find test -type f | while read file; do
    printf '%s\0%s\0%s\n' "${file%/*}" "$(tr -dc / <<< "$file")" "$file"
done | sort -t '\0' | awk -F'\0' '{print $3}'

It does the same thing, with ${file%/*} being used to get a file's directory name and the tr command being used to count the number of slashes, which is equivalent to a file's "depth".

(I sure hope there's an easier answer out there. What you're asking doesn't seem that hard, but I am blanking on a simple solution.)