Create a new folder named the next count - bash

Question

I have a folder of projects, which labels projects based on a number. It starts at 001 and continues counting. I have a bash script I run through Alfred, however, I currently have to type the name of the folder.
QUERY={query}
mkdir /Users/admin/Documents/projects/"$QUERY"
I would like to have the script automatically name the folder to the next number.
For example, if the newest folder is "019" then I would like it to automatically name it to "020"
This is what I've whipped up so far:
nextNum = $(find ~/documents/projects/* -maxdepth 1 -type d -print| wc -l)
numP = nextNum + 1
mkdir /Users/admin/Documents/projects/00"$numP"
I'm not sure if my variable syntax is correct, or if variables are the best way to do this. I am a complete noob to bash so any help is appreciated.

Answer 1

This might be what you're looking for:
#!/bin/bash
cd /Users/admin/Documents/projects || exit
if ! [[ -d 000 ]]; then mkdir 000; exit; fi
dirs=([0-9][0-9][0-9])
mkdir $(printf %03d $(( 1 + 10#${dirs[${#dirs[*]} - 1]} )) )

Answer 2

See comments in code to understand the code.
#!/bin/bash
# Configure your project dir
projects_dir=/Users/admin/Documents/projects
# Find the last project number.
# You can use wc instead of the sort|tail|sed I used. The result is the same.
last_project=$(\ls $projects_dir | sort -n | tail -n1 | sed 's/^0*//')
# Use printf to add the leading '0'
# ${last_project:-0} will substitute '0' when $last_project not set,
# therefore it will work even if your project directory is empty.
# $(( expr )) evaluates math expression. Also you can not have
# subshell expression in $(( expr )) so you can't substitute
# $last_project with the expression above
mkdir $projects_dir/$(printf "%03d" $((${last_project:-0} + 1)))
Although unlikely, the issue with the above script is that hitting a project count over 999 will break the 3 digits directory name convention. A simple fix is to use a 4 digits convention. A directory large amount of files/subdirectories is not recommended anyway so if you do reach 9999 projects, it's best to continue in a second project directory.

Answer 3

Give a try to this funny one:
projects_dir="/Users/admin/Documents/projects/"
[[ -n "${HOME}" ]] && projects_dir="${projects_dir/#~/$HOME}"
mkdir "${projects_dir}"/$(printf "%03d" $(find "${projects_dir}" -maxdepth 1 -type d -printf . | wc -c))
(EDIT)
Expanded version with bash trace output:
set -x
projects_dir="/Users/admin/Documents/projects/"
[[ -n "${HOME}" ]] && projects_dir="${projects_dir/#~/$HOME}"
cd "${projects_dir}" || exit 1
printf "each dot is a directory: "
find "${projects_dir}" -maxdepth 1 -type d -printf .
printf "\n"
let next_directory_id=$(find "${projects_dir}" -maxdepth 1 -type d -printf . | wc -c)
next_directory_name=$(printf "%03d" "${next_directory_id}")
mkdir "${projects_dir}"/"${next_directory_name}"
set +x

Answer 4

targ_number=$(find ~/documents/projects/ -maxdepth 1 -type d -print| wc -l)
mkdir /Users/admin/Documents/projects/$(printf "%03d" $targ_number)