List directory names using Ansible Find module - ansible

Question

I need to find folder names in /home/test directory. Once I have the list of folder names I need to loop over a task. I am able to retrieve the complete path, but not able to retrieve the folder name. is my approach wrong, or is there any easy way to retrieve folder names?
--
- hosts: localhost
gather_facts: no
vars:
Files: []
namespaces: []
tasks:
- name: Recursively find folders under /home/test
ansible.builtin.find:
paths: /home/test
file_type: directory
recurse: no
register: output
- name: Adding Files to the LIST
set_fact:
Files: "{{ Files + [item.path] }}"
with_items: "{{ output.files }}"
- name: Remove the path to only and retrieve folder name only
set_fact:
namespaces: "{{ namespaces + item | split('/') | last}}"
with_items: "{{ Files }}"
Error:
fatal: [localhost]: FAILED! => {"msg": "Unexpected templating type error occurred on ({{ namespaces + item | split('/') | last}}): can only concatenate list (not \"str\") to list"}

Answer 1

Put the below declaration into the vars
namespaces: "{{ output.files|map(attribute='path')|map('basename')|list }}"
Example of a complete playbook for testing
Given the tree
shell> tree /tmp/test/
/tmp/test/
├── dir_A
├── dir_B
├── file_1
└── file_2
the playbook
- hosts: localhost
vars:
namespaces: "{{ output.files|map(attribute='path')|map('basename')|list }}"
tasks:
- find:
paths: /tmp/test
file_type: directory
register: output
- debug:
var: namespaces
gives (abridged)
namespaces:
- dir_B
- dir_A

Answer 2

Your expression item | split('/') | last currently returns a string that you try to concatenate to the namespaces list. However, this is not allowed and thus you need to cast it to a list as well by simply put it into braces []:
- name: Remove the path to only and retrieve folder name only
set_fact:
namespaces: "{{namespaces + [item | split('/') | last]}}"
with_items: "{{ Files }}"