问题描述
我试图对以下表达式使用省略:
I tried to use omit with an expression like this:
id: "{{ openstack_networks.id | default(omit) }}"
但是，当未定义openstack_networks变量时，它似乎总是会因异常而失败.
But it seems that it keeps failing with an exception when openstack_networks variable is not defined.
编写此jinja2过滤器的正确方法是什么?
What is the correct way to write this jinja2 filter?
在openstack_networks.id不存在的情况下，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，但是100％有效的解决方案可以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) }}"
如果已定义openstack_networks但没有id键或未定义openstack_networks，则将为您提供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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