CODEWITHSUNDEEP
ansible
kiigass
kiigass

Reputation: 437

Ansible copy with bash command in path

I have an ansible task:

- name: Copy files from A to B
  copy:
    src: "{{ item }}"
    dst: /usr/lib/modules/$(uname -r)/misc/
  with_items:
   - file.one
   - file.two

The problem is, that ansible creates a directory called /usr/lib/modules/$(uname -r)/misc/ (literally) and copies the files there instead of substituting the $(uname -r) with the command's output.

How can I get this done?

Upvotes: 0

Views: 364

Answers (2)

mhutter
mhutter

Reputation: 2916

To add to @joppich's answer, if you want to substitute in a value that is not available as a fact, you can use Registered Variables:

- name: Determine kernel version
  command: uname -r
  register: r_uname

- name: Copy files from A to B
  copy:
    src: "{{ item }}"
    dst: /usr/lib/modules/{{ r_uname.stdout_lines[0] }}/misc/
  with_items:
   - file.one
   - file.two

Upvotes: 1

saj
saj

Reputation: 681

By default, ansible will run it's setup module and discover facts about the target system.

You can access these facts like any variable:

- name: Copy files from A to B
    copy:
  src: "{{ item }}"
  dst: /usr/lib/modules/{{ ansible_kernel }}/misc/
  with_items:
   - file.one
   - file.two

Upvotes: 4

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