Using if elif fi in shell scripts [duplicate]

Question

I'm not sure how to do an if with multiple tests in shell. I'm having trouble writing this script:

echo "You have provided the following arguments $arg1 $arg2 $arg3"
if [ "$arg1" = "$arg2" && "$arg1" != "$arg3" ]
then
    echo "Two of the provided args are equal."
    exit 3
elif [ $arg1 = $arg2 && $arg1 = $arg3 ]
then
    echo "All of the specified args are equal"
    exit 0
else
    echo "All of the specified args are different"
    exit 4
fi

The problem is I get this error every time:

./compare.sh: [: missing `]' command not found

Answer 1

Josh Lee's answer works, but you can use the "&&" operator for better readability like this:

echo "You have provided the following arguments $arg1 $arg2 $arg3" if [ "$arg1" = "$arg2" ] && [ "$arg1" != "$arg3" ] then      echo "Two of the provided args are equal."     exit 3 elif [ $arg1 = $arg2 ] && [ $arg1 = $arg3 ] then     echo "All of the specified args are equal"     exit 0 else     echo "All of the specified args are different"     exit 4  fi 

Answer 2

sh is interpreting the && as a shell operator. Change it to -a, that’s [’s conjunction operator:

[ "$arg1" = "$arg2" -a "$arg1" != "$arg3" ]

Also, you should always quote the variables, because [ gets confused when you leave off arguments.

Answer 3

Use double brackets...

if [[ expression ]]