如何为整个项目而不是单个项目创造条件 [英] How to have a condition for with-items as a whole, not for the individual items

问题描述

我有这个:

- name: Add hosts to /etc/hosts
  lineinfile:
    dest=/etc/hosts
    line='{{ item.dest }} {{ item.src }}'
    regexp='.*{{ item.src }}.*'
    state=present
  with_items:
    - "{{ hosts[service_name] }}"
  when: (service_name in hosts)

我得到的

'dict object' has no attribute u'blah'

如果when条件为false，我需要跳过整个任务.

What I need is to skip the whole task if the when condition is false.

推荐答案

使用with_items时，无法将when条件附加到整个任务上.

There is no way to attach when condition to a whole task when using with_items.

但是您可以采取另一种方式–如果没有密钥，则默认迭代器为空列表，如下所示:

But you can go another way – default iterator to empty list if there is no key, like this:

- name: Add hosts to /etc/hosts
  lineinfile:
    dest=/etc/hosts
    line='{{ item.dest }} {{ item.src }}'
    regexp='.*{{ item.src }}.*'
    state=present
  with_items: "{{ hosts[service_name] | default([]) }}"

在这种情况下，如果您提供未知的服务名称，则该任务将有零个要重复的项目.
还要注意，我在您的with_items结构中删除了不必要的列表定义.

In this case the task will have zero items to iterate over if you provide unknown service name.
Also note that I removed unnecessary list definition in your with_items construction.

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