仅从 ansible playbook 运行一项任务和处理程序 [英] Run only one task and handler from ansible playbook

问题描述

如何从 Ansible playbook 运行单个任务以及在该任务成功完成时收到通知的处理程序，同时跳过相关 playbook 中的所有其他任务?

How can I run a single task from an Ansible playbook and the handler that gets notified when that task completes successfully, while skipping all other tasks in the relevant playbook?

目前我执行以下操作:

ansible-playbook --start-at-task "task1" --step -K -i hosts playbook.yml

然后在任务完成后按Ctrl+c.但是，这也将跳过处理程序.

and then press Ctrl+c after the task has finished. This will also skip the handler however.

我知道我可以向任务添加标签并使用它，如 如何在 ansible playbook 中只运行一项任务?，但我希望能够在不添加标签的情况下执行此操作.这可能吗?

I know I can add a tag to the task and use that, as in How to run only one task in ansible playbook?, but I would prefer being able to do this without adding a tag. Is that possible?

推荐答案

目前 ansible-playbook 没有任何东西可以让您运行单个任务，例如 --task.因此，对我来说，标签和 --tags 选项是您最好的解决方案.

There's currently nothing coming with ansible-playbook to allow you to run a single task, like --task. Thus, to me, the tag along with the --tags option is your best solution here.

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