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Old 03-12-2009, 11:56 AM   #1
LQ Newbie
Registered: Mar 2009
Posts: 2

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how to skip 5 lines

how do you skip 5 lines from a text file and then search the text file for a specific word. after searching the word and display the each line corresponding to the search word.

so far i coded this much

echo -n "Enter a file name : "
read file

# make sure file exist
if [ ! -f $file ]
echo "$file not a file!"
exit 1

# put while loop to read a $file
while read line
#process each word
head 20 $file
for w in $line

Last edited by notoriouskingpin; 03-12-2009 at 11:58 AM.
Old 03-12-2009, 01:09 PM   #2
Registered: Jul 2007
Location: Scunthorpe, UK
Distribution: Ubuntu 8.10; Gentoo; Debian Lenny
Posts: 961

Rep: Reputation: 69
So you want to grep a file for a word, but don't want it to look in the first five lines? Well, there are a few ways though I'd reccommend python instead of bash. Python would be

#!/usr/bin/env python
#-*- encoding:utf-8 -*-

def worker(file, word):
        file = open( file ).readlines()[5:]
        print "File Does Not Exist"
        raise SystemExit

    for i in range( len( file) ):
        if word in file[i]:
            print file[i]

if __name__ == '__main__':
    from sys import argv
        worker( argv[1], argv[2] )
        print """ Syntax:
%s file word
""" % ( argv[0] )
If you're dead set with bash, you could get the number of lines (see man wc), minus 5 and then tail that into a variable. Then you could grep through that.
Old 03-12-2009, 02:09 PM   #3
LQ Veteran
Registered: Nov 2005
Location: Annapolis, MD
Distribution: Arch/XFCE
Posts: 17,802

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Is this homework??

I don't see anything about 5 lines in your code.

One thing to look at is SED addressing. You can specify a range of line numbers to search for a keyword.

Excellent SED tutorial here:


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