"unary operator expected" error in Bash if condition

Question

This script is getting an error:

elif [ $operation = "man" ]; then
    if [ $aug1 = "add" ]; then         # <- Line 75
    echo "Man Page for: add"
    echo ""
    echo "Syntax: add [number 1] [number 2]"
    echo ""
    echo "Description:"
    echo "Add two different numbers together."
    echo ""
    echo "Info:"
    echo "Added in v1.0"
    echo ""
elif [ -z $aug1 ]; then
    echo "Please specify a command to read the man page."
else
    echo "There is no manual page for that command."
fi

I get this error:

calc_1.2: line 75: [: =: unary operator expected

Answer 1

If you know you're always going to use Bash, it's much easier to always use the double bracket conditional compound command [[ ... ]], instead of the POSIX-compatible single bracket version [ ... ]. Inside a [[ ... ]] compound, word-splitting and pathname expansion are not applied to words, so you can rely on

if [[ $aug1 == "and" ]];

to compare the value of $aug1 with the string and.

If you use [ ... ], you always need to remember to double quote variables like this:

if [ "$aug1" = "and" ];

If you don't quote the variable expansion and the variable is undefined or empty, it vanishes from the scene of the crime, leaving only

if [ = "and" ];

which is not a valid syntax. (It would also fail with a different error message if $aug1 included white space or shell metacharacters.)

The modern [[ operator has lots of other nice features, including regular expression matching.

Answer 2

It took me a while to find this, but note that if you have a spacing error, you will also get the same error:

[: =: unary operator expected

Correct:

if [ "$APP_ENV" = "staging" ]

vs

if ["$APP_ENV" = "staging" ]

As always, setting -x debug variable helps to find these:

set -x

Answer 3

Try assigning a value to $aug1 before use it in if[] statements; the error message will disappear afterwards.

Answer 4

Make sure that when comparing a variable like $operation using string comparison you do not leave the possibility that $operation returns an empty string if you use the POSIX approach (single bracket).

Both [ $operation == "man"] and [ "" == "man" ] return the same error: [: ==: unary operator expected

Also, you shouldn't use the = operator unless you double bracket, as stated above.

Reference: https://tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

Answer 5

This error can also occur reading numerical input that could possibly be blank (to accept a default option).

The solution is to structure the if statement with multiple conditions & test for an empty variable first.

For example:

# sanitise input
var=$(echo $ans | tr -cd "[:digit:]")

if [ -z "$var" ] || [ "$var" -lt 1 ]; then
   do_something
fi

I had to solve this unary operator expected issue in remove_old_pkgs() of the helper script abk for Arch Sign Modules.

See also 6.4 Bash Conditional Expressions

Answer 6

You can also set a default value for the variable, so you don't need to use two "[", which amounts to two processes ("[" is actually a program) instead of one.

It goes by this syntax: ${VARIABLE:-default}.

The whole thing has to be thought in such a way that this "default" value is something distinct from a "valid" value/content.

If that's not possible for some reason you probably need to add a step like checking if there's a value at all, along the lines of "if [ -z $VARIABLE ] ; then echo "the variable needs to be filled"", or "if [ ! -z $VARIABLE ] ; then #everything is fine, proceed with the rest of the script".