conditional environment definition in Dockerfile [duplicate]

Question

Is it possible to conditionally set an ENV variable in a Dockerfile based on the value of a build ARG?

Ex: something like

ARG BUILDVAR=sad
ENV SOMEVAR=if $BUILDVAR -eq "SO"; then echo "hello"; else echo "world"; fi

Update: current usage based on Mario's answer:

ARG BUILD_ENV=prod
ENV NODE_ENV=production
RUN if [ "${BUILD_ENV}" = "test" ]; then export NODE_ENV=development; fi

However, running with --build-arg BUILD_ENV=test and then going onto the host, I still get

docker run -it mycontainer bin/bash
[root@brbqw1231 /]# echo $NODE_ENV
production

Answer 1

Yes, it is possible, but you need to use your build argument as flag. You can use parameter expansion feature of shell to check condition. Here is a proof-of-concept Docker file:

FROM debian:stable
ARG BUILD_DEVELOPMENT
# if --build-arg BUILD_DEVELOPMENT=1, set NODE_ENV to 'development' or set to null otherwise.
ENV NODE_ENV=${BUILD_DEVELOPMENT:+development}
# if NODE_ENV is null, set it to 'production' (or leave as is otherwise).
ENV NODE_ENV=${NODE_ENV:-production}

Testing build:

docker build --rm -t env_prod ./
...
docker run -it env_prod bash
root@2a2c93f80ad3:/# echo $NODE_ENV 
production
root@2a2c93f80ad3:/# exit
docker build --rm -t env_dev --build-arg BUILD_DEVELOPMENT=1 ./
...
docker run -it env_dev bash
root@2db6d7931f34:/# echo $NODE_ENV
development

Answer 2

You cannot run bash code in the Dockerfile directly, but you have to use the RUN command. So, for example, you can change ENV with RUN and export the variable in the if, like below:

ARG BUILDVAR=sad 
RUN if [ "$BUILDVAR" = "SO" ]; \
    then export SOMEVAR=hello; \
    else export SOMEVAR=world; \
    fi 

I didn't try it but should work.