CODEWITHSUNDEEP
bashif-statement
rahuL
rahuL

Reputation: 3420

Integer comparison in Bash using if-else

I have a variable called choice. Now, I try to use if to compare the entered value:

read $choice

if [ "$choice" ==  2  ];then
       #do something
elif [ "$choice" == 1 ];then
       #do something else
else
     echo "Invalid choice!!"
fi

The output goes directly to invalid choice if I enter either 1 or 2. I tried to put quotes around 1 and 2 inside the if statement. Still didn't work. Using -eq gives me an error "Unary operator expected".What am I doing wrong here?

Upvotes: 10

Views: 78356

Answers (3)

Mat
Mat

Reputation: 206679

Your read line is incorrect. Change it to:

read choice

i.e. use the name of the variable you want to set, not its value.

-eq is the correct test to compare integers. See man test for the descriptions (or man bash).

An alternative would be using arithmetic evaluation instead (but you still need the correct read statement):

read choice

if (( $choice == 2 )) ; then
    echo 2
elif (( $choice == 1 )) ; then
    echo 1
else
    echo "Invalid choice!!"
fi

Upvotes: 24

chepner
chepner

Reputation: 530960

An unpractical answer, just to point out that since read takes a variable name, the name can be specified by another variable.

choice=choice   # A variable whose value is the string "choice"
read $choice    # $choice expands to "choice", so read sets the value of that variable
if [[ $choice -eq 1 ]]; then   # or (( choice == 1 ))

Upvotes: 4

devnull
devnull

Reputation: 123458

For your example, case seems to be what you probably want:

read choice

case "$choice" in
"1")
    echo choice 1;
    ;;
"2")
    echo choice 2;
    ;;
*)
    echo "invalid choice"
    ;;
esac

Upvotes: 6

Related Questions