如果条件为真，则运行一些包含 yml 文件 [英] If condition is true run some include yml files

问题描述

我有一些 ubuntu 和 centos 的剧本，我想使用 main.yml 来检查 when: ansible_os_family =='RedHat' 或 ansible_distribution == 'Centos'，运行 playbooks(一些 ans many :-)).

I have some playbook for ubuntu and centos and I want to use main.yml to check when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos', run playbooks ( as some ans many :-) ).

当我跑步时:

-include: centos-xxx.yml
-include: centos-xaa.yml
-include: centos-xsss.yml

它将运行所有这些

基本上我希望剧本在满足条件时运行.

Basically I want that the playbook will run if meet condition.

我没有找到任何说明如何运行包含的文档:如果可能的话，我试图不为每个包含创建任务.

I didn't find any doc that say how to run include: more then one i am trying to not make task per include if possible.

推荐答案

您可以使用 有条件时 以包含文件.事实上，这有点常见.

You can use the when conditional to include files. This is somewhat common, in fact.

- include: centos-xxx.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'
- include: debian-xxx.yml
  when: ansible_distribution == 'Debian'

根据您的评论-如果您想按顺序运行它们，您有两个选择.这是直截了当的:

Per your comment- if you want to run them in order, you have two options. Here's the straightforward:

- include: centos-a.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'
- include: centos-b.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'
- include: centos-c.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'
- include: centos-d.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'

或者，您可以这样做:

- include: centos.yml
  when: ansible_os_family == 'RedHat' or ansible_distribution == 'Centos'

在 centos.yml 中:

- include: centos-a.yml
- include: centos-b.yml
- include: centos-c.yml
- include: centos-d.yml

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