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Top Forums Shell Programming and Scripting Printing previous line based on pattern using sed Post 302382537 by gaurav1086 on Wednesday 23rd of December 2009 02:38:20 PM
Old 12-23-2009
hello why not. this simple thing.
Code:
grep -B m -A n 'pattern' infile

m= number of lines you want before the pattern.
n= number of lines you want after the pattern.

Regards
 

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ZGREP(1)						      General Commands Manual							  ZGREP(1)

NAME
zgrep - search possibly compressed files for a regular expression SYNOPSIS
zgrep [ grep_options ] [ -e ] pattern filename... DESCRIPTION
Zgrep invokes grep on compressed or gzipped files. These grep options will cause zgrep to terminate with an error code: (-[drRzZ]|--di*|--exc*|--inc*|--rec*|--nu*). All other options specified are passed directly to grep. If no file is specified, then the standard input is decompressed if necessary and fed to grep. Otherwise the given files are uncompressed if necessary and fed to grep. If the GREP environment variable is set, zgrep uses it as the grep program to be invoked. EXIT CODE
2 - An option that is not supported was specified. AUTHOR
Charles Levert (charles@comm.polymtl.ca) SEE ALSO
grep(1), gzexe(1), gzip(1), zdiff(1), zforce(1), zmore(1), znew(1) ZGREP(1)
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