Split string to get all but the last element

userfRKK 0
Viewed 62

Here is my playbook

---
- name: This is a hello-world example
  hosts: localhost
  vars:
    mystr: "I.To.Be.war"
  tasks:

    - name: Hi
      debug:
        msg: "{{ mystr.split('.') | first | trim }}"

Output:

I

The trim filter is to get rid of \r\n at the end of the variable incase it has.

However, I wish a generic solution that should print everything before the last element of the split delimiter, i.e.: ..

Expected output:

I.To.Be

Note: I do not want a solutions like "{{ mystr.split('.')[0][1][2] }}" as it is not generic in nature.

formatting ansible split
3 Answers
  1. Apply the trim filter to every element of your list using the map filter
  2. Using python list slicing to get everything but the last
  3. Join the elements back
$ ansible localhost -m debug \
-a "msg={{ (mystr.split('.') | map('trim'))[:-1] | join('.') }}" \
-e "mystr=I.to.be.war"
localhost | SUCCESS => {
    "msg": "I.to.be"
}

Alternatively, you can use a regex to match the first part of the string

$ ansible localhost -m debug \
-a 'msg={{ mystr | regex_replace("^(.*)\\.[^.]*$", "\\1") }}' \
-e "mystr=I.to.be.war"
localhost | SUCCESS => {
    "msg": "I.to.be"
}
Zeitounator0

Since your requirement seems to be about getting the name of a file without its extension, mind that there is a extra Jinja filter, splitext, purposed for that:

Where, the task

- debug:
    var: "'I.To.Be.war' | splitext | first | trim"

will give your expected

ok: [localhost] => 
  '''I.To.Be.war'' | splitext | first | trim': I.To.Be

While the filter splitext is actually returning a tuple, more convenient than a regular split:

ok: [localhost] => 
  '''I.To.Be.war'' | splitext | trim': ('I.To.Be', '.war')
β.εηοιτ.βε0

You could use Ansible's regex_replace() filter:

mystr | regex_replace('\\.[^.]+$', '')
Allan Wind0
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