Bash Scripting General Questions (conditionals and variable passing)

Question

I'm rather new to bash scripting, and Google isn't as useful as I'd like for this. I'm just playing around with a little password entry program in my .bash_profile and have something like this:

read PASSWORD
if $PASSWORD != 'pass'; then
    echo "wrong. exiting"
    exit
fi

Unfortunately, this doesn't work. I get these errors (darwin on 10.6)...

EDIT Sorry about this posting. My browser crashed and I didn't even realize this posted. I ended up figuring it out on my own – again sorry. But thanks for the answers!

Answer 1

Try:

read PASSWORD
if [ "x$PASSWORD" != "xpass" ] ; then
   echo "Wrong. Exiting."
   exit 1
fi
exit 0

Answer 2

You are missing square brackets. The if line should be:

if [ $PASSWORD != 'pass' ]; then

or even better:

if [ "$PASSWORD" != 'pass' ]; then

Which will avoid failure if $PASSWORD is empty.

Answer 3

You might like to know about two options to the read command:

-p string

Display a prompt without a trailing newline

and

-s

Silent mode. The characters typed by the user are not echoed to the screen.

So for prompting for a password you could do:

read -sp "Please enter your password: " PASSWORD
echo

This is an excellent resource.

Answer 4

use case/esac construct

read -p "enter: " PASSWORD
case "$PASSWORD" in
    "pass") echo "ok;;
    * ) echo "not ok";;
esac

Edit: For Dennis's qns

x=10
y=5
z=1
options=3
expression="$((x> y)) $((y> z)) $((options<=4))"
case "$expression" in
   "1 1 1")
    echo "x > y and y>z and options <=4";;
    *) echo "Not valid";;
esac