Remove first line in 'stdout_lines' - automation

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.