如何删除或排除Ansible模板列表中的项目？ [英] How to remove or exclude an item in an Ansible template list?

问题描述

我正在编写一个Ansible模板，该模板需要在主机组中生成IP列表， 不包括 。我已经在网上和文档中进行了搜索，但是找不到任何允许您删除列表中项目的过滤器。我在下面创建了（hacky）for循环来执行此操作，但我想知道是否有人知道这样的最佳实践过滤方法。

I'm writing an Ansible template that needs to produce a list of ip's in a host group, excluding the current hosts IP. I've searched around online and through the documentation but I could not find any filters that allow you to remove an item in a list. I have created the (hacky) for loop below to do this but was wondering if anyone knew a "best practice" way of filtering like this.

{% set filtered_list = [] %}

{% for host in groups['my_group'] if host != ansible_host %}
    {{ filtered_list.append(host)}}
{% endfor %}

可以说groups ['my_group']有3 ip（192.168.1.1、192.168.1.2和192.168.1.3）。当为192.168.1.1生成模板时，应该只打印IP的192.168.1.2和192.168.1.3。

Lets say groups['my_group'] has 3 ip's (192.168.1.1, 192.168.1.2 and 192.168.1.3). When the template is generated for 192.168.1.1 it should only print the ip's 192.168.1.2 and 192.168.1.3.

推荐答案

差异 过滤器为此：

- debug: var=item
  with_items: "{{ groups['my_group'] | difference([inventory_hostname]) }}"

这将为您提供 my_group 没有当前主机。

This will give you all items hosts from my_group without current host.

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