在 Ansible 角色内执行特定任务

Question

如何执行某些任务。例如对于以下角色，假设我正在部署应用程序，那么我只需要创建通用目录和 Application1 目录。我怎样才能只部署这两个任务？

我有以下 Ansible 目录结构：

├── ansible.cfg
├── playbook.yml
└── roles
  └── groups
    └── tasks
      └── main.yml

我的 groups/tasks/main.yml 文件如下所示：

---
- name: Create Directory General
  file:
    path: /tmp/general
    state: directory
    mode: '0755'

- name: Create Directory Application1
  file:
    path: /tmp/Application1
    state: directory
    mode: '0755'

- name: Create Directory Application2
  file:
    path: /tmp/Application2
    state: directory
    mode: '0755'

Playbook.yml

---
- name: Deploy application
  hosts: localhost
  become: yes
  become_method: sudo
  become_user: root
  roles:
    - groups

注意：我已经知道 include_role 但它会为角色内的所有任务设置一个标签。我不想那样做。

Answer 1

IMO 对这个问题没有一个好的答案，但既然你提到了tags...

---
- name: Create Directory General
  file:
    path: /tmp/general
    state: directory
    mode: '0755'
  tags:
    - always

- name: Create Directory Application1
  file:
    path: /tmp/Application1
    state: directory
    mode: '0755'
  tags:
    - app1

- name: Create Directory Application2
  file:
    path: /tmp/Application2
    state: directory
    mode: '0755'
  tags:
    - app2

然后你可以调用你的剧本：

# Deploy all apps with common tasks
ansible-playbook playbook.yml

# Deploy only application 1 with common tasks
ansible-playbook --tags app1 playbook.yml

# Deploy application 2 without running common tasks
ansible-playbook --tags app2 --skip-tags always playbook.yml

请注意最新的，因为它也会默认跳过事实收集。有关详细信息，请参阅上面的文档链接。