How to determine what containers use the docker volume? - docker

Question

Suppose I have a volume and I know its name or id.
I want to determine the list of containers (their names or ids) that use the volume.
What commands can I use to retrieve this information?
I thought it can be stored in the output of docker volume inspect <id> command but it gives me nothing useful other than the mount point ("/var/lib/docker/volumes/<id>").

Answer 1

docker ps can filter by volume to show all of the containers that mount a given volume:
docker ps -a --filter volume=VOLUME_NAME_OR_MOUNT_POINT
Reference: https://docs.docker.com/engine/reference/commandline/ps/#filtering

Answer 2

The script below will show each volume with the container(s) using it:
#!/bin/sh
volumes=$(docker volume ls --format '{{.Name}}')
for volume in $volumes
do
echo $volume
docker ps -a --filter volume="$volume" --format '{{.Names}}' | sed 's/^/ /'
done
Listing volumes by container is slightly trickier so it's an exercise for the reader but the above should suffice unless you have many containers/volumes.

Answer 3

This is related to jwodder suggestion, if of any help to someone.
It basically gives the summary of all the volumes, in case you have more than a couple and are not sure, which is which.
import io
import subprocess
import pandas as pd
results = subprocess.run('docker volume ls', capture_output=True, text=True)
df = pd.read_csv(io.StringIO(results.stdout),
encoding='utf8',
sep=" ",
engine='python')
for i, row in df.iterrows():
print(i, row['VOLUME NAME'])
print('-' * 20)
cmd = ['docker', 'ps', '-a', '--filter', f'volume={row["VOLUME NAME"]}']
print(subprocess.run(cmd,
capture_output=True, text=True).stdout)
print()

Answer 4

For each volume, this script outputs the list of containers using this volume, bearing in mind that a volume may be used by several containers.
for v in $(docker volume ls --format "{{.Name}}")
do
containers="$(docker ps -a --filter volume=$v --format '{{.Names}}' | tr '\n' ',')"
echo "volume $v is used by $containers"
done