Split string to get all but the last element

Question

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.

Answer 1

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"
}

Answer 2

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')