"unary operator expected" error in Bash if condition

Question

I've been trying to figure out whats wrong with this but just can't figure it out..

This is the part seems to be 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.