问题描述
在终端中逐行运行命令时，以下工作正常：
The following works fine when running the commands manually line by line in the terminal:
docker create -it --name test path
docker start test
docker exec test /bin/sh -c "go test ./..."
docker stop test
docker rm -test
但是当我作为一个shell脚本运行时，Docker容器既不停止也不被删除。
But when I run it as a shell script, the Docker container is neither stopped nor removed.
#!/usr/bin/env bash
set -e
docker create -it --name test path
docker start test
docker exec test /bin/sh -c "go test ./..."
docker stop test
docker rm -test
如何使其在shell脚本中工作？
How can I make it work from within a shell script?
推荐答案
如果您使用 set -e ，则在任何命令失败时，脚本将退出。即当命令返回代码！= 0时，这意味着如果您的开始， exec 或停止失败，您将留下一个集装箱。
If you use set -e the script will exit when any command fails. i.e. when a commands return code != 0. This means if your start, exec or stop fails, you will be left with a container still there.
您可以删除 set -e ，但您可能仍然希望使用 go test 命令作为总体返回码。
You can remove the set -e but you probably still want to use the return code for the go test command as the overall return code.
#!/usr/bin/env bash
docker create -it --name test path
docker start test
docker exec test /bin/sh -c "go test ./..."
rc=$?
docker stop test
docker rm test
exit $rc
陷阱
使用 set -e 实际上是非常有用的，并且捕获了许多在大多数脚本中被忽略的问题。一个稍微复杂的解决方案是在 EXIT ，这意味着 set -e 可以使用。
Trap
Using set -e is actually quite useful and catches a lot of issues that are silently ignored in most scripts. A slightly more complex solution is to use a trap to run your clean up steps on EXIT, which means set -e can be used.
#!/usr/bin/env bash
set -e
# Set a default return code
RC=2
# Cleanup
function cleanup {
echo "Removing container"
(docker stop test)
(docker rm -f test)
exit $RC
}
trap cleanup EXIT
# Test steps
docker create -it --name test path
docker start test
docker exec test /bin/sh -c "go test ./..."
RC=$?
这篇关于在shell脚本中执行时，Docker kill不工作的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！