我是 Ansible 的新手,并试图获得一个服务状态,其中服务名称是动态的,并且在 playbook 之前由 set_fact 设置。
如何在另一个变量中构建一个变量? 我希望我可以使用类似的东西来显示我的服务状态:
{{ ansible_facts.services['{{ servicename }}'].state }}
但是它不起作用。
所以我用 vars 尝试了这种方式:
- name: set service name
ansible.builtin.set_fact:
servicename: "'myservice'"
when: ansible_distribution_major_version == "7"
- name: print service state
debug: msg={{ vars['ansible_facts.services[' + servicename + '].state'] }}
vars:
servicename: "{{ servicename }}"
我收到以下错误:
fatal: [myhost]: FAILED! => {"msg": "The task includes an option with an undefined variable. The error was: 'dict object' has no attribute u\"ansible_facts.services['myservice'].state\"\n\nThe error appears to be in '/etc/ansible/playbook/myplaybook.yaml': line 20, column 5, but may\nbe elsewhere in the file depending on the exact syntax problem.\n\nThe offending line appears to be:\n\n\n - name: print service state\n ^ here\n"}
当以下工作正常时:
- name: print service state
debug:
msg: "{{ ansible_facts.services['myservice'].state }}"
答案 0 :(得分:1)
在 {{ … }}
块中,您可以直接访问变量。从我的角度来看,您可以省略 ansible_facts
以及第二个任务中的变量赋值。
这会做你想做的事(正如 Zeitounator 已经写过的那样):
- hosts: localhost
vars:
services:
myservice:
state: foo
tasks:
- name: set service name
ansible.builtin.set_fact:
servicename: "myservice"
- name: print service state
debug:
msg: "{{ services[servicename].state }}"