Split parameter and use it as multiple arguments

I'm trying to create a reusable role that does a synchronize where I pass in the src, dest and an exclude parameters.

I'd like to split that exclude parameter into multiple --exclude= parameters if it's defined.

So, effectively what would be running is:

With exclude undefined:

- name: Copy extracted files to remote
  synchronize:
    src: "{{ src }}"
    dest: "{{ dest }}"
    rsync_opts:
      - "--no-motd"

With exclude=logs,conf:

- name: Copy extracted files to remote
  synchronize:
    src: "{{ src }}"
    dest: "{{ dest }}"
    rsync_opts:
      - "--no-motd"
      - "--exclude=logs"
      - "--exclude=conf"

I figure I need to split the variable, but I'm not sure how to add the --exclude= to each one, whilst not adding any if there's nothing passed.

Solution

Given that you are passing the exclude parameter to your role via:

roles:
  - role: demo_role
    src: /path/to/src
    dest: /path/to/dest
    exclude:
      - logs
      - conf

Then, in your synchronize task, you could use a product filter, in order to construct the exclusions path along with the --exclude flag:

rsync_opts: >-
  {{
    ['--no-motd']
    + ['--exclude=']
      | product(exclude | default([]))
      | map('join')
  }}

From there:

when exclude is a list:

exclude:
   - logs
   - conf

rsync_opts will be:

- --no-motd
- --exclude=logs
- --exclude=conf

when exclude is an empty list:
```
exclude: []
```
rsync_opts will be:
```
- --no-motd
```
when exclude is undefined:
```
- role: demo_role
  src: /path/to/src
  dest: /path/to/dest
```
rsync_opts will be:
```
- --no-motd
```
thanks to the usage of the default filter