Docker exec in bash script

Question

I'm creating dynamically docker containers via bash script:

while getopts ":s:d:h" opt; do
    case $opt in
        s)
            for i in $(seq $2 $END); 
                do 
                    docker run -dit --name=app_client_$i -d app:client
                    docker exec -d app_client_$i $app_start
            done
            ;;
...

The docker container starts fine, but the docker exec command caused problems. When I try (without the -d):

docker exec app_client_$i $app_start

The application inside the docker container starts fine - but I'm attached to this docker container. I want to start the app inside the docker container in the background, so I use the -d parameter:

docker exec -d app_client_$i $app_start

With that the app doesn't start inside the docker container. What I am missing?

I've already tried that. I get nothing from the logs. If I type docker logs app_client_$1 there is no output. — jwi
– jwi, Commented Feb 26, 2016 at 10:15
The value of $app_start is a java call with the java interpreter. — jwi
– jwi, Commented Feb 26, 2016 at 10:15
Weird thing in your docker run -itd which should be either a daemon or interactive, but not both, and docker exec should not be with -d, See supervisor, runit, daemontools or s6 you want to launch several processes in a container — user2915097
– user2915097, Commented Feb 26, 2016 at 10:19
Okay, tried it without the -i parameter inside the run command and also tried it with the quoting variables like "app_client_$i" "$app_start". Both do not work. — jwi
– jwi, Commented Feb 26, 2016 at 10:26

jwi · Accepted Answer · 2016-02-26 10:42:38Z

2

Okay, got it (facepalm):

With the docker -d you're going to start the process INSIDE the container in the background. So my app was already running inside the container, but in the background.

Cheers!

answered Feb 26, 2016 at 10:42

jwi

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