Ansible: Extract first three octects from an IP address

Question

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?

You want `192.168.10` to be returned? – helloV Nov 29 '16 at 01:17 — helloV, Nov 29 '16 at 01:17

score 12 · Accepted Answer · answered Nov 29 '16 at 06:07

12

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

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

answered Nov 29 '16 at 06:07

Konstantin Suvorov

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this is a cool trick and clear. So in general can we use any python methods through pipe in ansible? – spamulap12 Nov 29 '16 at 17:26
No, pipes are for filters, see my [other](http://stackoverflow.com/a/39794185/2795592) answer for details. But you can use python methods of strings and dicts with dot-notation. – Konstantin Suvorov Nov 29 '16 at 17:54

Ibrahim · Answer 2 · 2016-11-29T03:22:25.937

2

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')}}

edited Nov 29 '16 at 03:22

answered Nov 29 '16 at 01:23

Ibrahim

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thanks, but it has some issue with syntax I bileve. The offending line appears to be: set_fact: net: {{comp_ip | regex_replace('^([0-9]{1,3}.[0-9]{1,3}.[0-9]{1,3})'), '\\1'}} ^ here – spamulap12 Nov 29 '16 at 01:30
No syntax error with this, but o/p has nothing in it {"net": "(u'', '\\\\1')"} – spamulap12 Nov 29 '16 at 01:42
@spamulap12 Try the updated full code, I made a small change. By the way, does the input has only `ip`? or does it have something before, and after the `ip`? – Ibrahim Nov 29 '16 at 02:02
This is the full ansible code I am trying with `code` - name: COMP IP set_fact: comp_ip: "192.168.10.2" - name: Extract net set_fact: net: "{{comp_ip | regex_replace('^[\s\S]*?([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})[\s\S]*$'), '\\1'}}" I get the syntax error like before :( – spamulap12 Nov 29 '16 at 02:09
The full code has a typo, the closing bracket of regex_replace should be after the \\1 `{{comp_ip | regex_replace('^([0-9]{1,3}\.[0-9]{1,3}\.[0-9]{1,3})\.[0-9]{1,3}$', '\\1')}}` – spamulap12 Nov 29 '16 at 03:20
@spamulap12 Yes, you're right, how did we miss that. – Ibrahim Nov 29 '16 at 03:23
@spamulap12 my pleasure. – Ibrahim Nov 29 '16 at 03:24

score 1 · Answer 3 · answered Feb 05 '18 at 12:14

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') }}"

Ansible: Extract first three octects from an IP address

3 Answers3