UNIX for Dummies Questions & Answers

#if there are no arguements (IE -r, -l, ect) than output the file
if [ $# -eq 0 ]
then
cat < dirInfo

#if the first arguement is -n, output any line that contains the 2nd argument (this WORKS, and if the name is the 2nd argument it will output the whole line of information)
elif [ "$1" = "-n" ]
then
grep $2 dirInfo

#I keep getting a "syntax error near unexpected token `('" error.  I figured I could use the grep for the phone number just like the name?  
#I have also tried something like grep '[0-9]\{3\}-[0-9]\{4\}'	but dont know how to compare with 2nd arguement.
elif [ "$1" == "-p" ]
then
grep $2 dirInfo

#I cant grep a variable? This seems to be one of the main things stopping me, I dont know how to incorporate the variable (their input) into grep.
elif [ "$1" == "-l" ]
then
grep '^$2' dirInfo

#As for finding an area code, I would assume thats very similar to finding the whole number, with a small edit of the code.
elif [ "$1" == "-c" ]
then

#Appending a name that they enter (arguement 2).  This does not work at all.
elif [ "$1" == "-a" ]
then
sed '$a\$2'

#=: command not found
elif [ "$1" == "-r" ]
then
echo "Enter the new name: "
read name
echo "Enter the new address: "
read address
echo "Enter the new phone number: "
read phonenumber
$line = grep $2 dirInfo
sed 's/$line/name, address, phonenumber/p' dirInfo
else
echo "not found."
fi

grep "$2" dirInfo

grep "$2"'$' file

#Doesnt Work
elif [ "$1" == "-p" ]
then
grep "$2" dirInfo

./phonebook -p "(324)343-3421"