根据条件在ansible中传递Shell命令

问题描述

我对Ansible非常陌生，在此方面面临一些困难 我必须基于ansible中的某些条件传递完整的shell命令 我试图设置事实，但是我无法执行shell并由于未定义变量而出错

msg“：”该任务包括带有未定义变量的选项。错误是：“ commands_to_run”未定义\ n \ n

- name: setting command for x
    set_fact:
     commands_to_run: ls -1d /filepath/* | grep app | grep  -E "("{{ id }}".*){2}"
    when: id  == 'X'

  - name:setting command for Y
    set_fact:
     commands_to_run: ls -1d /filepath/* | grep app| grep -e "{{ id }}"
    when: id  != 'X'

  - name: list path
    shell: "{{ commands_to_run }}"
    register: path

需要根据条件执行的Shell命令完全不同。 请让我知道是否可以通过其他方式完成

ansible-playbook 2.4.2.0

预先感谢

解决方法

修复您的代码标识，以下示例有效：

文件 site.yml ：

---
- hosts: localhost
  connection: local
  gather_facts: false
  tasks:
  - name: setting command for X
    set_fact:
      commands_to_run: "echo Case X"
    when: id == 'X'
  - name: setting command for Y
    set_fact:
      commands_to_run: "echo Case Y"
    when: id != 'X'
  - name: call command
    shell: "{{ commands_to_run }}"
    register: path

呼叫案例X：ansible-playbook site.yml -e id = X -v 通话案例Y：ansible-playbook site.yml -e id = Y -v

如果必须将案例Y称为默认案例，则可以设置默认值：

- hosts: localhost
  connection: local
  gather_facts: false
  tasks:
  - name: setting command for Y
    set_fact:
      commands_to_run: "echo Case Y"
  - name: setting command for X
    set_fact:
      commands_to_run: "echo Case X"
    when: id == 'X'
  - name: call command
    shell: "{{ commands_to_run }}"
    register: path

如果无法定义id，请使用默认值：

- hosts: localhost
  connection: local
  gather_facts: false
  tasks:
  - name: setting command for Y
    set_fact:
      commands_to_run: "echo Case Y"
  - name: setting command for X
    set_fact:
      commands_to_run: "echo Case X"
    when: id | default('Y') == 'X'
  - name: call command
    shell: "{{ commands_to_run }}"
    register: path