How to use docker container's env variable while running docker exec command

Question

I am running a docker exec -it ... command and I need to use an environment variable of my docker container. An example:

docker exec -it container_id command_here param_1 $param_2_as_env_variable

In the case above, it pulls param_2_as_env_variable from host machine, and not the docker container. Is it possible to use env variable from container itself while running docker exec ... command from another machine?

Update: I can use ouput of docker exec -it container_id printenv | grep .... But I couldn't separate value and key. How can I get only value here?

Henry · Accepted Answer · 2020-07-06 08:15:33Z

4

Something like this could work (it assumes, the container has a shell installed)

docker exec -it container_id sh -c 'command_here param_1 $param_2_as_env_variable'

For example the following works:

docker exec -it test sh -c 'echo $HOSTNAME'

to give the host name of the container.

edited Jul 6, 2020 at 8:15

answered Jul 6, 2020 at 8:03

Henry

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4 Comments

JollyRoger Over a year ago

Nope, this still gets host machine's env variable.

JollyRoger Over a year ago

Now it accepts as a plain string. Doesn't parse at all :(

Henry Over a year ago

Are you sure the environment variable is defined in the container?

JollyRoger Over a year ago

Yes, I can see it when I run printenv. I updated the question, can you check?

arco444 · Accepted Answer · 2020-07-06 08:03:14Z

3

You need to escape the variable to pass it to the docker exec command unresolved:

Try:

docker exec -it container_id command_here param_1 \$param_2_as_env_variable

answered Jul 6, 2020 at 8:03

arco444

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3 Comments

JollyRoger Over a year ago

This produces illegal input exception. So it doesn't parse $param2 at all, it just gets it as plain string.

KamilCuk Over a year ago

This produces illegal input exception. So it doesn't parse How does "illegal input exception" (what does "produce" the message exactly? What is the message exactly?) lead to "doesn't parse"?

JollyRoger Over a year ago

Sorry, the error wasn't related this case. The answer is correct.

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