'Stopping Docker containers by image name - Ubuntu
On Ubuntu 14.04 (Trusty Tahr) I'm looking for a way to stop a running container and the only information I have is the image name that was used in the Docker run command.
Is there a command to find all the matching running containers that match that image name and stop them?
Solution 1:[1]
The previous answers did not work for me, but this did:
docker stop $(docker ps -q --filter ancestor=<image-name> )
Solution 2:[2]
You could start the container setting a container name:
docker run -d --name <container-name> <image-name>
The same image could be used to spin up multiple containers, so this is a good way to start a container. Then you could use this container-name to stop, attach... the container:
docker exec -it <container-name> bash
docker stop <container-name>
docker rm <container-name>
Solution 3:[3]
This code will stop all containers with the image centos:6. I couldn't find an easier solution for that.
docker ps | grep centos:6 | awk '{print $1}' | xargs docker stop
Or even shorter:
docker stop $(docker ps -a | grep centos:6 | awk '{print $1}')
Solution 4:[4]
Two ways to stop running a container:
1. $docker stop container_ID
2. $docker kill container_ID
You can get running containers using the following command:
$docker ps
Following links for more information:
Solution 5:[5]
Stop docker container by image name:
imagename='mydockerimage'
docker stop $(docker ps | awk '{split($2,image,":"); print $1, image[1]}' | awk -v image=$imagename '$2 == image {print $1}')
Stop docker container by image name and tag:
imagename='mydockerimage:latest'
docker stop $(docker ps | awk -v image=$imagename '$2 == image {print $1}')
If you created the image, you can add a label to it and filter running containers by label
docker ps -q --filter "label=image=$image"
Unreliable methods
docker ps -a -q --filter ancestor=<image-name>
docker ps -a -q --filter="name=<containerName>"
filters by container name, not image name
docker ps | grep <image-name> | awk '{print $1}'
is problematic since the image name may appear in other columns for other images
Solution 6:[6]
I made a /usr/local/bin/docker.stop that takes in the image name (assumes you only have one running).
docker stop $(docker ps -q -f "name=$1")
Solution 7:[7]
list all containers with info and ID
docker ps
docker stop CONTAINER ID
Solution 8:[8]
For Docker version 18.09.0 I found that format flag won't be needed
docker rm $(docker stop $(docker ps -a -q -f ancestor=<image-name>))
Solution 9:[9]
I was trying to wrap my Docker commands in gulp tasks and realised that you can do the following:
docker stop container-name
docker rm container-name
This might not work for scenarios where you have multiple containers with the same name (if that's possible), but for my use case it was perfect.
Solution 10:[10]
In my case --filter ancestor=<image-name> was not working, so the following command cleaned up the Docker container for me:
docker rm $(docker stop $(docker ps -a -q --filter "name=container_name_here" --format="{{.ID}}"))
Solution 11:[11]
Adding on top of @VonC superb answer, here is a ZSH function that you can add into your .zshrc file:
dockstop() {
docker rm $(docker stop $(docker ps -a -q --filter ancestor="$1" --format="{{.ID}}"))
}
Then in your command line, simply do dockstop myImageName and it will stop and remove all containers that were started from an image called myImageName.
Solution 12:[12]
use: docker container stop $(docker container ls -q --filter ancestor=mongo)
(base) :~ user$ docker ps
CONTAINER ID IMAGE COMMAND CREATED STATUS PORTS NAMES
d394144acf3a mongo "docker-entrypoint.s…" 15 seconds ago Up 14 seconds 0.0.0.0:27017->27017/tcp magical_nobel
(base) :~ user$ docker container stop $(docker container ls -q --filter ancestor=mongo)
d394144acf3a
(base) :~ user$ docker ps
CONTAINER ID IMAGE COMMAND CREATED STATUS PORTS NAMES
(base) :~ user$
Solution 13:[13]
This is my script to rebuild docker container, stop and start it again
docker pull [registry]/[image]:latest
docker build --no-cache -t [localregistry]/[localimagename]:latest -f compose.yaml context/
docker ps --no-trunc | grep [localimagename] | awk '{print $1}' | xargs docker stop
docker run -d -p 8111:80 [localregistry]/[localimagename]:latest
note --no-trunc argument which shows the image name or other info in full lenght in the output
Solution 14:[14]
Here's a concise command which doesn't require you to specify the image tag (as most of these answers do):
docker stop $(docker ps -a | awk -v i="^${image_name}.*" '{if($2~i){print$1}}')
Solution 15:[15]
docker stop $(docker ps -a | grep "zalenium")
docker rm $(docker ps -a | grep "zalenium")
This should be enough.
Solution 16:[16]
If you want to prefer a simple AWK approach, here Is my take:
docker rm -f $(docker ps | awk '{ if($2 == "<your image name>") { print $NF}}')
$(docker ps | awk '{ if($2 == "<your image name>") { print $NF}}') - prints the docker container names based on input image
docker ps - list all containers
awk '{ if($2 == "<your-image-name>") { print $NF}}' - The second parsed column of docker ps gives the image name. Comparing it with your image name will execute print $NF which prints the container name.
docker rm -f removes the containers
For example, removing all running containers of ubuntu image, can be done simply as:
docker rm -f $(docker ps | awk '{ if($2 == "ubuntu:latest") { print $NF}}')
PS: Remember to include the image tag in AWK, since it's a equal comparator.
Solution 17:[17]
if you know a part of the container name you can use AWK with docker as following :
CONTAINER_IDS=$(docker ps -a | awk '($2 ~ /container.*/) {print $1}')
if [ -z "$CONTAINER_IDS" -o "$CONTAINER_IDS" == " " ]; then
echo "No containers available for deletion"
else
docker rm -f $CONTAINER_IDS
fi
Solution 18:[18]
This will only stop all containers with image = "yourImgName" :
sudo docker stop $(sudo docker ps | grep "yourImgName" | cut -d " " -f 1)
This will stop and remove all containers with image = "yourImgName" :
sudo docker rm $(sudo docker stop $(sudo docker ps -a | grep "yourImgName" | cut -d " " -f 1))
Sources
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Source: Stack Overflow