How to limit `docker run` execution time?

问题:

I want to run a command inside a docker container. If the command takes more than 3 seconds to finish, the container should be deleted.

I thought I can achieve this goal by using --stop-timeout option in docker run.

But it looks like something goes wrong with my command.

For example, docker run -d --stop-timeout 3 ubuntu:14.04 sleep 100 command creates a docker container that lasts for more than 3 seconds. The container is not stopped or deleted after the 3rd second.

Do I misunderstand the meaning of --stop-timeout?

The document says

--stop-timeout Timeout (in seconds) to stop a container

Here's my docker version:

Client:
 Version:       17.12.0-ce
 API version:   1.35
 Go version:    go1.9.2
 Git commit:    c97c6d6
 Built: Wed Dec 27 20:03:51 2017
 OS/Arch:       darwin/amd64

Server:
 Engine:
  Version:      17.12.0-ce
  API version:  1.35 (minimum version 1.12)
  Go version:   go1.9.2
  Git commit:   c97c6d6
  Built:        Wed Dec 27 20:12:29 2017
  OS/Arch:      linux/amd64
  Experimental: true

The API version is newer than 1.25.

回答1:

The --stop-timeout option is the maximum amount of time docker should wait for your container to stop when using the docker stop command.

A container will stop when it's told to or when the command is running finishes, so if you change you sleep from 100 to 1, you'll see that the container is stopped after a second.

What I'll advice you to do is to change the entrypoint of your container to a script that you create, that will execute what you want and keep track of the execution time from within and exit when timeout.

After that you can start your container using the --rm option that will delete it once the script finishes.

A small example.

Dockerfile:

FROM ubuntu:16.04

ADD ./script.sh /script.sh

ENTRYPOINT /script.sh

script.sh:

#!/bin/bash

timeout=5
sleep_for=1

sleep 100 &

find_process=$(ps aux | grep -v "grep" | grep "sleep")

while [ ! -z "$find_process" ]; do
    find_process=$(ps aux | grep -v "grep" | grep "sleep")

    if [ "$timeout" -le "0" ]; then
      echo "Timeout"
      exit 1
    fi

    timeout=$(($timeout - $sleep_for))
    sleep $sleep_for
done

exit 0

Ran it using:

docker build -t testing .
docker run --rm testing

This script will execute sleep 100 in background, check if its still running and if the timeout of 5 seconds is reach then exit.

This might not be the best way to do it, but if you want to do something simple it may help.

回答2:

You can try

timeout 3 docker run...

there is a PR on that subject

https://github.com/moby/moby/issues/1905

See also

Docker timeout for container?