CODEWITHSUNDEEP
regexansiblejinja2
spamulap12
spamulap12

Reputation: 97

Ansible: Extract first three octects from an IP address

I have a string with IP addr: 192.168.10.2

I want to extract first three octets of the IP in Ansible and I tried to use this regex. 

{{comp_ip | regex_replace("^[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}"), "//1"}}

This does not yield any result. Can someone correct me where I went wrong?

Upvotes: 3

Views: 9263

Answers (3)

panticz
panticz

Reputation: 2315

In case you want to calculate your network address you can use Ansible ipaddr filter which provides exactly this functionality: http://docs.ansible.com/ansible/latest/playbooks_filters_ipaddr.html

---
- hosts: localhost
  vars:
    my_ip: "{{ ansible_default_ipv4.network }}/{{ ansible_default_ipv4.netmask }}"
  tasks:
    - debug: msg="network {{ my_ip | ipaddr('network') }}"
    - debug: msg="netmask {{ my_ip | ipaddr('netmask') }}"

Upvotes: 1

Konstantin Suvorov
Konstantin Suvorov

Reputation: 68339

If already have dot-separated IP-address, there is a simple way:

{{ comp_ip.split('.')[0:3] | join('.') }}

Upvotes: 12

Ibrahim
Ibrahim

Reputation: 6098

You are doing it right, you just have to use parenthesis in Regex to make a group. It is better to match the whole ip and end your regex with $, and also change //1 to \\1 in your code.

Change regex from:

^[0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3}

To this regex:

^([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})\.[0-9]{1,3}$

This is the full code:

{{comp_ip | regex_replace('^([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})\.[0-9]{1,3}$', '\\1')}}

Upvotes: 2

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