如何在一行/语句中检查bash中的退出代码？

Question

这就是我想要做的:

#!/bin/bash
set -e # I can't remove this line!

if [ docker inspect foo ]; then
  echo container is present
else
  echo container is absent
fi

可能吗？当容器不存在以及何时存在时,docker inspect foo返回退出代码.10

现在我明白了:

-bash: [: too many arguments

Answer 1

如果要在"if"语句中运行该命令,则可以这样执行:

if docker inspect foo
then
  echo container is present
else
  echo container is absent
fi

如果需要,可以省略所有换行符或用分号替换.这也有效:

if docker inspect foo; then echo container is present; else echo container is absent; fi

如果你想要更紧凑的东西,你可以使用这种语法:

docker inspect foo && echo container is present
docker inspect foo || echo container is absent
docker inspect foo && echo container is present || echo container is absent

&&如果第一个命令成功,则运行第二个命令.||如果第一个命令失败,则运行第二个命令.最后一行使用两种形式.