Get files from directories alphabetically sorted with bash - bash

Question

I have this code that works in the directory that I execute:
pathtrabajo=.
filedirs="files.txt"
dirs=$(find . -maxdepth 1 -mindepth 1 -type d | sort -n)
SAVEIFS=$IFS
IFS=$(echo -en "\n\b")
for entry in $dirs
do
echo "${entry}" >> "${filedirs}"
find "$entry" -maxdepth 1 -mindepth 1 -name '*.md' -printf '%f\n' | sort | sed 's/\.md$//1' | awk '{print "- [["$0"]]"}' >> "${filedirs}"
done
IFS=$SAVEIFS
But when I try to make it global to work with variables, find gives error:
pathtrabajo="/path/to/a/files"
filedirs="files.txt"
dirs=$(find "${pathtrabajo}" -maxdepth 1 -mindepth 1 -type d | sort -n)
SAVEIFS=$IFS
IFS=$(echo -en "\n\b")
for entry in "${dirs[#]}"
do
echo "${entry}" >> "${pathtrabajo}"/"${filedirs}"
find "${entry}" -maxdepth 1 -mindepth 1 -name '*.md' -printf '%f\n' | sort | sed 's/\.md$//1' | awk '{print "- [["$0"]]"}' >> "${pathtrabajo}"/"${filedirs}"
done
IFS=$SAVEIFS
What did I do wrong?

Answer 1

It's really not clear why you are using find here at all. The following will probably do what you are trying if I can guess from your code.
dirs=([0-9][!0-9]*/ [0-9][0-9][!0-9]*/ [0-9][0-9][0-9][!0-9]*/ [!0-9]*/)
printf "%s\n" "${dirs[#]}" >"$filedirs"
for dir in "${dirs[#]}"; do
printf "%s\n" "$dir"/*.md |
awk '{ sub(/\.md$/, ""); print "- [["$0"]]" }'
done >>"$filedirs"
The shell already expands wildcards alphabetically. The dirs assignment will expand all directories which start with a single digit, then the ones with two digits, then the ones with three digits -- extend if you need more digits -- then the ones which do not start with a digit.
It would not be hard, but cumbersome, to extend the code to run in an arbitrary directory. My proposed solution would be (for once!) to cd to the directory where you want the listing, then run this script.