如果Ansibe中的变量为空，如何省略? [英] How to omit if variable is empty in Ansibe?

问题描述

我正在研究 Docker 的 ansible 示例.

I'm going through ansible example for Docker.

我需要像示例中一样发布一个端口，但使用 if 语句 if port else omit.像这样:

And I need to publish a port just like in the example but with an if statement if port else omit. Like this:

docker_container:
name: myapplication
...
ports:
 - "{{ if port else omit }}" # the {{ port }} variable is set from the default task.
...

但是每次我运行它时，Docker 守护进程都会告诉我:

But each time I run this, the Docker daemon tells me:

template error while templating string: expected token 'end of print statement', got 'port'. String: {{ if port else omit }}"

如果变量 {{ port }} 为空，我如何省略设置端口?

How do I omit setting the port if the variable {{ port }} is empty?

提前感谢您的帮助.

推荐答案

您应该使用 default 过滤器，带有 omit 变量.

You should use the default filter with the omit variable for that.

- docker_container:
    name: myapplication
    ports: "{{ port|default(omit) if port is not defined else [port] }}"

请注意，您要使用这种确切的语法，而不是像

Mind that you want to use this exact syntax and not the array notation like

- docker_container:
    name: myapplication
    ports:
      - "{{ port|default(omit) }}"

否则你将在属性 ports 中得到一个数组，但仍然会有一些奇怪的值，比如

Otherwise you will end up with an array in the attribute ports, still, that would have some odd value like

"__omit_place_holder__ad3616ee8afa39aa187d7fc6ac7ad36f3e7691c0"

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