Extracting part of the string using Ansible regex_search and save the output as a variable

Question

I'm having a content like below inside a file.
dataDir=/var/lib/zookeeper
4lw.commands.whitelist=mntr,conf,ruok,stat
syncLimit=2

I wanted to read the value for dataDir using Ansible and set it to a variable. I have written following code but regular expression and storing the variable both have some issues.
  - name: Read zoo.cfg content
    shell:
      cmd: cat zoo.cfg
    register: zoo_config_content

- set_fact:
      my_var: "{{ zoo_config_content.stdout | regex_search('dataDir.*')}}"

- name: Print
    debug:
      var: my_var

Q1.) How can I improve the regular expression to get only /var/lib/zookeeper ?
Q2.) How can I store that extracted value to use in another task?

Gerald Schneider · Accepted Answer

You need to add a group to your regex and a second parameter that specifies which group to return:
- set_fact:
    my_var: "{{ zoo_config_content.stdout | regex_search('dataDir=(.+)', '\1') | first }}"

This would return an array with a single element, so I added | first to get only one element directly as a string.
Result:
ok: [localhost] => {
    "my_var": "/var/lib/zookeeper"
}

If the config file resides on the Ansible machine you can make it even easier using a lookup:
- set_fact:
    my_var: "{{ lookup('file', 'zoo.cfg') | regex_search('dataDir=(.+)', '\1') | first }}"

Extracting part of the string using Ansible regex_search and save the output as a variable

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