How to sort the results of find (including nested directories) alphabetically in bash - 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/file
test/file
test/z/file
I want to sort the output so it appears as:
test/file
test/a/file
test/b/file
test/z/file
Is there any way to sort the results within the find command, or by piping the results into sort?

Answer 1

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.)

Answer 2

find test -type f -printf '%h\0%p\n' | sort | awk -F'\0' '{print $2}'
The result of find is, for example,
test/a'\0'test/a/file
test'\0'test/file
test/z'\0'test/z/file
test/b'\0'test/b/text file.txt
test/b'\0'test/b/file
where '\0' stands for null character.
These compound strings can be properly sorted with a simple sort:
test'\0'test/file
test/a'\0'test/a/file
test/b'\0'test/b/file
test/b'\0'test/b/text file.txt
test/z'\0'test/z/file
And the final result is
test/file
test/a/file
test/b/file
test/b/text file.txt
test/z/file
(Based on the John Kugelman's answer, with "depth" element removed which is absolutely redundant.)

Answer 3

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 4

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.