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Top Forums Shell Programming and Scripting Returning only part of a line when grepping Post 302410255 by riott on Monday 5th of April 2010 12:24:55 PM
Old 04-05-2010
Returning only part of a line when grepping

I want to grep out a part of a snort rule based on the SID given, but all i want as the output is the part in the quotes after the msg: An example line looks something like this:

alert tcp any any -> 127.0.0.1 any (msg:"Example Message"; classtype:Example; sid:123456;)

I would want it to only show Example Message.

Any help is appreciated.

Thanks!

~Riott
 

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ZGREP(1)						    BSD General Commands Manual 						  ZGREP(1)

NAME
zgrep, zegrep, zfgrep -- print lines matching a pattern in gzip-compressed files SYNOPSIS
zgrep [grep-flags] [--] pattern [files ...] zegrep [grep-flags] [--] pattern [file ...] zfgrep [grep-flags] [--] pattern [file ...] DESCRIPTION
zgrep runs grep(1) on files or stdin, if no files argument is given, after decompressing them with zcat(1). The grep-flags and pattern arguments are passed on to grep(1). If an -e flag is found in the grep-flags, zgrep will not look for a pattern argument. zegrep calls egrep(1), while zfgrep calls fgrep(1). EXIT STATUS
In case of missing arguments or missing pattern, 1 will be returned, otherwise 0. SEE ALSO
egrep(1), fgrep(1), grep(1), gzip(1), zcat(1) AUTHORS
Thomas Klausner <wiz@NetBSD.org> BSD
December 28, 2003 BSD
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