Remove first line in 'stdout_lines'

Question

I'm just a beginner and write some small playbook.

For now I get some little error.

- name: Facts
  ios_command:
    commands: show interface status
  register: ios_comm_result

In this playbook, for beginner I get the list of interfaces and then in next task I get list with first position including column name.

- name: get a list of ports to modify
  set_fact:
    newlist: "{{ (newlist | default([])) + [item.split()[0]] }}"
  with_items: "{{ ios_comm_result.stdout_lines }}"

Output:

'Port'
'fa0/1'
'gi0/2'
....
 etc

How can I delete first line? Or where can I read about this?

Answer 1

Based on your requirement and the given comments about Accessing list elements with Slice notation ([1:]) and Jinja Filters (| reject()) I've setup an short test

---
- hosts: localhost
  become: false
  gather_facts: false

  vars:

    STDOUT_LINES: ['Port', 'fa0/1', 'gi0/2', 'etc/3']

  tasks:

  # Slice notation

  - name: Show 'stdout_lines' from the 2nd line to the end
    debug:
      msg: "{{ item }}"
    loop: "{{ STDOUT_LINES[1:] }}"

  # Jinja2 filter

  - name: Show 'stdout_lines' and drop any lines which match
    debug:
      msg: "{{ item }}"
    loop: "{{ STDOUT_LINES | reject('match', '^Port') | list }}"

  # Conditionals

  - name: Show 'stdout_lines' but not the first one
    debug:
      msg: "{{ item }}"
    when: ansible_loop.index != 1
    loop: "{{ STDOUT_LINES }}"
    loop_control:
      extended: true
      label: "{{ ansible_loop.index0 }}"

with some value added about Extended loop variables and using conditionals in loops.

All three options show the expected result.