'How to turn a variable into a dictionary item
I am struggling with what I believe is the simplest of tasks. I want to take a variable v and turn that into a dictionary where
v: ValueofV
The end goal is to write that dictionary to a JSON file that contains a list of variables, with key: value where key name is always a variable name, so for variables a, b, c, ... I should end up with:
{
"a": "a_val",
"b": "v_val",
"c": "c_val"
}
I've tried building lists with_items, e.g.
- name: Var3
set_fact:
node_state4: "{{ node_state4 | default({}) | combine({ item : item })}}"
with_items:
- requested_node_count
- added_node_count
But, that makes the value the string name. If I make the second item {{ item }} it fails.
Solution 1:[1]
Use filter community.general.dict_kv e.g.
_dict: "{{ v|community.general.dict_kv('v') }}"
gives
_dict:
v: ValueofV
Given the list of the variables
rnodes: [a, b, c]
a: a_val
b: b_val
c: c_val
iterate the list and create the dictionary, e.g.
- set_fact:
_dict: "{{ _dict|d({})|
combine(lookup('vars', item)|
community.general.dict_kv(item)) }}"
loop: "{{ rnodes }}"
gives
_dict:
a: a_val
b: b_val
c: c_val
The next option is to extract the variables and use filters dict and zip, e.g. the task below gives the same result
- set_fact:
_dict: "{{ dict(rnodes|zip(_vals)) }}"
vars:
_vals: "{{ rnodes|map('extract', vars) }}"
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 |
