下面是一个脚本，用于删除创建时间超过2天的正在运行和已退出的容器:

#!/bin/bash
# This script will kill and remove containers older than 2 days.
#
docker ps -aq > /scripts/containers.txt
today=`date +%Y-%m-%d`
oldate=`date --date="2 day ago" +%Y-%m-%d`
while read p; do
    cont=`docker inspect -f '{{ .Created }}' $p | cut -c 1-10`
    echo " Created date of $p is $cont"
    k=`echo $(( ( $(date -ud $today +'%s') - $(date -ud $cont +'%s'))/60/60/24 ))`
    echo $k
    if [ $k -ge 2 ];
    then
            echo "Killing docker container $p"
            docker kill $p
            echo "Removing docker container $p"
            docker rm $p
    else
            echo "Docker container $p is not one day old, so keeping the container."
    fi
done </scripts/containers.txt

移除所有容器:

sudo docker ps -a | grep -v CONTAINER | awk '{print $1}' | xargs --no-run-if-empty sudo docker rm -f

解释:

sudo docker ps -a

返回一个容器列表。

awk '{print $1}'

获取容器ID的第一列。

grep -v CONTAINER

删除标题。

最后一个管道将id安全地发送到sudo docker rm -f。

这个简短的脚本可能会有帮助(从以前的答案编译)!

#!/bin/bash

# Remove dangling images

IMAGE_IDS=$(sudo docker images -f "dangling=true" -q)

if [ -n "$IMAGE_IDS" ]; then

    sudo docker rmi $IMAGE_IDS > /dev/null 2>&1
    echo 'Images removed' $IMAGE_IDS
fi


# Remove exited containers

CONTAINER_IDS=$(sudo docker ps -a -q -f status=exited)

if [ -n "$CONTAINER_IDS" ]; then

    sudo docker rm -v $CONTAINER_IDS > /dev/null 2>&1
    echo 'Containers remove $CONTAINER_IDS'
fi

删除从某个映像创建的所有容器:

docker rm  $(docker ps -a | awk '/myimage:mytag/{print $1}')

删除所有docker进程: Docker rm $(Docker ps -a -q) 移除特定容器: $ docker ps -a列出所有旧容器 $ docker rm container-Id

Bash的这两行代码会在删除容器之前通过一些关键字来过滤它们:

containers_to_keep=$(docker ps -a | grep 'keep\|Up\|registry:latest\|nexus' | awk '{ print $1 }')
containers_to_delete=$(docker ps -a | grep Exited | grep -Fv "$containers_to_keep" | awk '{ print $1 }')
docker rm $containers_to_delete

来自这篇文章。