Error in Shell script

sreedivia · #1 (**permalink**) 10-21-08, 22:28

Hi,
I want to write a shell script in that i am sending a file name as argument.
I want to check whether the file contain "#student layout" . if so i have to skip next 3 lines and print the line after that.
FILE="$1"
key=" "
i=0
exec 3<&0
exec 0<$FILE
while read line
do
echo $line
if [ $line == "#student layout" ]
then
for i in 1 2 3
do
read line
done
fi
done
exec 0<&3

exit 0

Can anyone tell me is there any syntaxt error in this. I am getting an error "syntax error near unexpected token 'do in line no 12.
I am working on Solaris

pdreyer · #2 (**permalink**) 10-22-08, 02:57

Syntax seem OK
your problem is that you do not quote your variable resulting in something funny on your if statement
change to
if [ "$line" ...etc.

LKBrwn_DBA · #3 (**permalink**) 10-22-08, 07:54

This will work on some "Unix" flavors:

Code:

grep -N 4 "#student layout" $FILE|tail -1

mike_bike_kite · #4 (**permalink**) 10-22-08, 08:37

And just to prove there's loads a ways to skin a cat:

Program:

Code:

#!/bin/sh

echo -n "File:"
read file

cat $file | sed -n '/#student layout/{n;n;n;n;p;}'

exit 0

test data:

Code:

# cat tmp.dat
line 1
#student layout
line 3
line 4
line 5
line 6

#student layout
line 9
line 10
line 11
line 12
line 13
#student layout

Output:

Code:

# ./tmp
File:tmp.dat
line 6
line 12

Internet Services

Web Development

Digital Marketing

Consumer Tech