How can escape colon in a string within an Ansible YAML file?

生来就可爱ヽ(ⅴ<●) 提交于 2019-11-28 08:07:14

you need to enclose the entire line in ", where : appears.

lineinfile:
'dest=/var/www/kibana/config.js
backrefs=true
regexp="(elasticsearch.* \"http.*)$"
line="elasticsearch\: \ {{ elasticsearch_URL }}:{{ elasticsearch_port }} \ "
state=present'  

See these pages:
Link-1 Link-2 Link-3

The solution that will work in any case no matter how many nested quotes you might have and without forcing you to add more quotes around the whole thing (which can get tricky to impossible depending on the line you want to write) is to output the colon through a Jinja2 expression, which simply returns the colon as a string:

{{ ":" }}

Or in your complete line:

line="elasticsearch\: \" {{ elasticsearch_URL }}{{ ":" }}{{ elasticsearch_port }} \" "

Credit to this goes to github user drewp.

Just keep the colon in quotes separately -

regexp="(elasticsearch.* \"http.*)$" line="elasticsearch':' \" {{ elasticsearch_URL }}:{{ elasticsearch_port }} \" "

foo=bar is the more suitable format for a one-line directive, but as you're already spanning several lines with your parameters anyway, just change the = to :, and it won't fuss about having a colon in your string.

- name: Comment out elasticsearch the config.js to ElasticSearch server
  lineinfile:
    dest:     /var/www/kibana/config.js
    backrefs: true
    regexp:   'elasticsearch.* "http.*$'
    line:     'elasticsearch: "{{ elasticsearch_URL }}:{{ elasticsearch_port }}"'
    state:    present

It’s a string already; you don’t have to (and can’t, as seen here) escape colons inside it.

line="elasticsearch: \" {{ elasticsearch_URL }}:{{ elasticsearch_port }} \" "
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