Shell Programming and Scripting

when i run this script:

# Author: xtos

if [ "$#" -ne 1 ]
then
echo "Please enter ONLY one name to search, (example "$0" NAME)"
exit
fi

INPUT="grep $1 $HOME/work/phone_book"

if [ $INPUT !="$1" ]

then

echo "Listing not found, please try again."

else


NAME=" `$INPUT | cut -f1 -d":"` "
PHONE=" `$INPUT | cut -f2 -d":"` "


echo $NAME
echo $PHONE

fi




i get this when the argument is right:

$ phone work
./phone: line 11: [: too many arguments
work
(312) 675-3321

why do i get too many arguments??

and when the argument is wrong i am supposed to get this :

"Listing not found, please try again."


but i dont, and instead i get this :


$ phone wooork
./phone: line 11: [: too many arguments

line 11 is :

if [ $INPUT !="$1" ]



what am i doing wrong?? any help would be much appreciated

thanks

Xtos,
If the '$1' value is not found in the file '$HOME/work/phone_book', the
result is empty:
INPUT="grep $1 $HOME/work/phone_book"

Then when you issue the 'if' statement:
if [ $INPUT !="$1" ]

The shell converts the content of 'INPUT' and the 'if' becomes:
if [ !="$1" ]

Thus, giving you the error.

If you want to know if a string is present, here is one way:

Code:

mCnt=`grep -c $1 $HOME/work/phone_book`
if [ $mCnt -eq 0 ]; then
  echo "Listing not found, please try again."
fi

grep returns 0 if the pattern is found and 1 if not.

Code:

#!/bin/ksh
if [ "$#" -ne 1 ]
then
        echo "Please enter ONLY one name to search, (example "$0" NAME)"
        exit
fi

grep $1 $HOME/work/phone_book > /dev/null

if [ $? -eq 1 ]
then
        echo "Listing not found, please try again."
else
        NAME=" `grep $1 $HOME/work/phone_book| cut -f1 -d":"` "
        PHONE=" `grep $1 $HOME/work/phone_book| cut -f2 -d":"` "
        echo $NAME
        echo $PHONE
fi