Bash conditionals: how to “and” expressions? (if [ ! -z $VAR && -e $VAR ])

Question

I guess I\'m not clear on how to do \"and\" tests. I wanted to make sure an argument existed which was working well with [ -e $VAR ], but it turns out that was also

Answer 1

From the bash manpage:

[[ expression ]] - return a status of 0 or 1 depending on the evaluation of the conditional expression expression.

And, for expressions, one of the options is:

expression1 && expression2 - true if both expression1 and expression2 are true.

So you can and them together as follows (-n is the opposite of -z so we can get rid of the !):

if [[ -n "$var" && -e "$var" ]] ; then
    echo "'$var' is non-empty and the file exists"
fi

However, I don't think it's needed in this case, -e xyzzy is true if the xyzzy file exists and can quite easily handle empty strings. If that's what you want then you don't actually need the -z non-empty check:

pax> VAR=xyzzy
pax> if [[ -e $VAR ]] ; then echo yes ; fi
pax> VAR=/tmp
pax> if [[ -e $VAR ]] ; then echo yes ; fi
yes

In other words, just use:

if [[ -e "$var" ]] ; then
    echo "'$var' exists"
fi