if statement - Integer comparison in Bash using if-else -

i have variable called choice. now, try use if compare entered value:

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

the output goes directly invalid choice if enter either 1 or 2. tried put quotes around 1 , 2 inside if statement. still didn't work. using -eq gives me error "unary operator expected".what doing wrong here?

your read line incorrect. change to:

read choice 

i.e. use name of variable want set, not value.

-eq correct test compare integers. see man test descriptions (or man bash).

an alternative using arithmetic evaluation instead (but still need correct read statement):

read choice  if (( $choice == 2 )) ;     echo 2 elif (( $choice == 1 )) ;     echo 1 else     echo "invalid choice!!" fi