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Asked: 2016-09-16 02:43:31 +0800 CST

How to do regex-replace in a file using Ansible?

  • 772

Based on this example:

- lineinfile: dest=/opt/jboss-as/bin/standalone.conf regexp='^(.*)Xms(\d+)m(.*)$' line='\1Xms${xms}m\3' backrefs=yes

from this documentation, it was tried to do a regex-replace in Ansible.

Ansible version

user@server:/home$ ansible --version
ansible 2.1.1.0

/path/to/file:

helloworld

Ansible snippets:

- lineinfile:
  dest: /path/to/file
  regexp='^(hello)world$'
  line='\1030'

attempt 2

- lineinfile:
  dest: /path/to/file
  regexp='^(hello)world$'
  line="\1030"

Expected outcome:

hello030

Current outcome:

\1030

Questions

  1. Why is the outcome \1030 instead of hello030?
  2. How to solve it?
ansible

2 Answers

  1. Best Answer

    Why is the outcome \1030 instead of hello030?

    The lineinfile module default is backrefs: false. Your regexp='^(hello)world$' matches the entire contents of file. Literal from line='\1030' replaces contents.

    How to solve it?

    1. Turn on backrefs with backrefs: true
    2. Use a named group in line:

    A backref followed by numbers will not function as expected. You will need a named group instead. e.g. \g<1>

    - name: Replace the world
  lineinfile:
    dest: file
    regexp: '^(hello)world$'
    line: '\g<1>030'
    backrefs: true
    • 11

  2. I suppose it is because it matches whole \1030 (as 1030-th backref). Maybe try \1 030 first and you will see, if it is the reason.

    • -2