How to check exit code in bash in one line/statement?

Question

This is what I'm trying to do:

#!/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

Is it possible? docker inspect foo returns exit code 1 when container is absent and 0 when it's present.

Now I'm getting this:

-bash: [: too many arguments

Answer 1

If you want to run the command within the "if" statement, you'd do it like this:

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

All of the line breaks can be omitted or replaced by semicolons if you want. This would also work:

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

If you want something more compact, you could use this kind of syntax:

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

&& runs the second command if the first succeeded. || runs the second command if the first one failed. The last line uses both forms.