Question

I need to understand bash if expressions. Sure, I’ve used Google. I know that it goes something like this:

if [ expr operator expr ]
then
doSomeThing
fi

However, I understand that Bash doesn’t have a boolean data type. I want to check if the file passed as an argument ($1) exists. The way I would do this straight of my mind:

if [ -e $1 = true ] 
then 
echo "File exists."
fḯ

Or maybe like:

if [ -e $1 ] #Assuming that this is true only if file in $1 exists. 

None of these work, and I’m not sure what [ ] means. -e $1 seems like a smart choice, but it is always true? There are different operators for strings and integers. And I can’t use parenthesis to group together expressions. This is so confusing.

Anyone got a few hints? The IF in bash does not work like if I’ve tried in any other language.

Answer 1

[ … ]

means that the program /usr/bin/test gets executed with ... as the arguments and its return value is checked (0 means true and x != 0 means false (yes, 0 really means true here because 0 is the OK exit code in UNIX)). So

if [ -e $1 ]; then
    echo "ok"
fi

is correct. It’s the same as

if test -e $1; then
    echo "ok"
fi

The sole problem is that $1 could contain spaces. If so /usr/bin/test gets confused, since it gets 3 or more arguments. The first one (-e) is a unary operator (wants one argument). Since you gave it 2 or more (one /usr/bin/test argument is the -e itself), it complains like that:

...: binary operator expected

So, simply use

if [ -e "$1" ];then
    echo "ok"
fi

That will even work if $1 contains spaces. Another possibility is to use

if [[ -e $1 ]]; then
    echo "ok"
fi

That will do the same but it gets evaluated by bash itself and no /usr/bin/test program gets forked.

Compounds are as usual but -a means and and -o means or. So

if [ -e /etc/fstab -a -e "$1" ]; then
   echo "ok"
fi

will echo ok if /etc/fstab and the file given as first command line argument exist.