How to get test command return code in the shell?

Question

Is it any better way to get return code from command in one line. eg:

$ test $(ls -l) ||echo 'ok'
-bash: test: too many arguments
ok

the above script have error in test command, because it seems parsing the output "ls - l" not return code.

I know use the "if" syntax is work fine, But need more then one lines.

ls -l
if [ $? -eq 0 ];then
   echo 'ok'
fi

Answer 1

You can use && and || to make these things one-liner. For example, in the following:

ls -l && echo ok

echo ok will run only if the command before && (ls -l) returned 0.

On the other hand, in the following:

ls -l || echo 'not ok'

echo 'not ok' will run only if the command before || returned non zero.

Also, you can make your if..else block one-liner using ;:

if ls -l;then echo ok;else echo 'not ok';fi

But this may make your code hard to read, so not recommended.

Answer 2

The if statement is catching the return value of a command, for example with ls:

if ls -l; then
    echo 'ok'
fi