bash Docker Run 和变量替换
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Docker Run and variable substitution
提问by Abimbola Esuruoso
I have a variable in a script called image_name
我在一个名为的脚本中有一个变量 image_name
image_name="my-app"
And I am attempting to pass this variable to a docker container as seen below:
我正在尝试将此变量传递给 docker 容器，如下所示：
docker build -t $image_name .
docker run --rm --name $image_name -e "app_dir=${image_name}" $image_name
But this variable is not set and is blank within the context of the docker build
但是这个变量没有设置并且在 docker build 的上下文中是空白的
This is a snippet from the Dockerfile below:
这是来自以下 Dockerfile 的片段：
....
RUN echo dir is $app_dir
....
This is a snippet of the build output below:
这是以下构建输出的片段：
....
Step 2 : RUN echo dir is $app_dir
---> Running in db93a939d701
dir is
---> c9f5e2a657d5
Removing intermediate container db93a939d701
....
Anyone know how to do the variable substitution?
有人知道如何进行变量替换吗？
采纳答案by Luís Bianchin
Notice that you are using the variable inside the Dockerfile. That variable will be evaluated at build time (docker build), so you need to pass its value at that moment. That can be achieved with the --build-argparam.
请注意，您正在使用Dockerfile. 该变量将在构建时 ( docker build)进行评估，因此您需要在那时传递其值。这可以通过--build-arg参数实现。
Changing your example, it would be:
更改您的示例，它将是：
docker build --build-arg app_dir=${image_name} -t $image_name .
docker run --rm --name $image_name $image_name
回答by Abimbola Esuruoso
This is achievable using the build-argparameter of the docker buildcommand
这可以使用命令的build-arg参数来实现docker build
See: https://github.com/docker/docker/issues/6822#issuecomment-168170031for an example
参见：https: //github.com/docker/docker/issues/6822#issuecomment-168170031示例
