问题描述
我尝试将 omit 与这样的表达式一起使用:
I tried to use omit with an expression like this:
id: "{{ openstack_networks.id | default(omit) }}"
但是当 openstack_networks
变量未定义时,它似乎一直失败并出现异常.
But it seems that it keeps failing with an exception when openstack_networks
variable is not defined.
编写这个 jinja2 过滤器的正确方法是什么?
What is the correct way to write this jinja2 filter?
如果 openstack_networks.id 不存在,我想省略该参数.
I want to omit the parameter in case openstack_networks.id does not exists.
推荐答案
不是超级优雅,但 100% 工作解决方案来处理可能未定义的父字典的键:
Not super elegant, but 100% working solution to handle keys of possibly undefined parent dicts:
id: "{{ (openstack_networks | default({})).id | default(omit) }}"
如果 openstack_networks
已定义但没有 id
键,或者 openstack_networks
未定义,这将为您提供 omit
.
This will give you omit
if openstack_networks
is defined but has no id
key or if openstack_networks
is undefined.
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