how to skip 5 lines

notoriouskingpin · 03-12-2009, 10:56 AM

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

#!/bin/sh
echo -n "Enter a file name : "
read file
jones=0

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

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

jamescondron · 03-12-2009, 12:09 PM

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

Code:

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

def worker(file, word):
    try:
        file = open( file ).readlines()[5:]
    except:
        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
    try:
        worker( argv[1], argv[2] )
    except:
        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.

pixellany · 03-12-2009, 01:09 PM

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:
http://www.grymoire.com/Unix/Sed.html