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Top Forums Shell Programming and Scripting Print selection of line based on line number Post 302394827 by mohanm on Friday 12th of February 2010 06:59:40 PM
Old 02-12-2010
Print selection of line based on line number

Hi Unix gurus

Basically i am searching for the pattern and getting the line numbers of the grepped pattern. I am trying to print the series of lines from 7 lines before the grepped line number to the grepped line number.

I am trying to use the following code. but it is not working.
Code:
cat filename | grep -n pattern | sed 's/:/ /g' | awk 'NR==$1-7,NR==$1'

Please help

Last edited by Scott; 02-12-2010 at 08:24 PM.. Reason: Please use code tags
 

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BWILD(8)						     Network backup, utilities							  BWILD(8)

NAME
bwild - Bacula's 'wildcard' engine SYNOPSIS
bwild [options] -f <data-file> DESCRIPTION
This manual page documents briefly the bwild command. This is a simple program that will allow you to test wild-card expressions against a file of data. OPTIONS
A summary of options is included below. -? Show version and usage of program. -d nn Set debug level to nn. -dt Print timestamp in debug output -f <data-file> The data-file is a filename that contains lines of data to be matched (or not) against one or more patterns. When the program is run, it will prompt you for a wild-card pattern, then apply it one line at a time against the data in the file. Each line that matches will be printed preceded by its line number. You will then be prompted again for another pattern. Enter an empty line for a pattern to terminate the program. You can print only lines that do not match by using the -n option, and you can suppress printing of line numbers with the -l option. -n Print lines that do not match -l Suppress lines numbers. -i use case insensitive match. SEE ALSO
fnmatch(3) AUTHOR
This manual page was written by Bruno Friedmann <bruno@ioda-net.ch>. Kern Sibbald 30 October 2011 BWILD(8)
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