How to write Unix shell scripts with options?

I don't know whether it's possible, but I want to write shell scripts that act like regular executables with options. As a very simple example, consider a shell script foo.sh that is configured to be executable:

    ./foo.sh
    ./foo.sh -o

and the code foo.sh works like

    #!/bin/sh
    if  ## option -o is turned on
        ## do something
    else
        ## do something different
    endif

Is it possible and how to do that? Thanks.

$ cat stack.sh 
#!/bin/sh
if  [[ $1 = "-o" ]]; then
    echo "Option -o turned on"
else
    echo "You did not use option -o"
fi

$ bash stack.sh -o
Option -o turned on

$ bash stack.sh
You did not use option -o

FYI:

$1 = First positional parameter
$2 = Second positional parameter
.. = ..
$n = n th positional parameter

For more neat/flexible options, read this other thread: Using getopts to process long and short command line options