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Top Forums Shell Programming and Scripting Printing a block of lines from a file, if that block does not contain two patterns using sed Post 302747427 by Kesavan on Friday 21st of December 2012 09:14:10 AM
Old 12-21-2012
Quote:
Originally Posted by Scrutinizer
awk OK? try something like this:
Code:
awk 'p{p=p RS $0} /\.\.\.\./ {if(p~s && !(p~/stop/)){print p; p=x}else p=$0}' s="12.12.2012" file

Hello Scrutinizer,

Thanks for the reply.
This awk statement will not work, if the date is different for example
if date is "11.12.2012".

I want to print the block, which does not has pattern "stop" and does not has pattern "13.12.2012" in the real time it will be the current date.
 

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IGAWK(1)							 Utility Commands							  IGAWK(1)

NAME
igawk - gawk with include files SYNOPSIS
igawk [ all gawk options ] -f program-file [ -- ] file ... igawk [ all gawk options ] [ -- ] program-text file ... DESCRIPTION
Igawk is a simple shell script that adds the ability to have ``include files'' to gawk(1). AWK programs for igawk are the same as for gawk, except that, in addition, you may have lines like @include getopt.awk in your program to include the file getopt.awk from either the current directory or one of the other directories in the search path. OPTIONS
See gawk(1) for a full description of the AWK language and the options that gawk supports. EXAMPLES
cat << EOF > test.awk @include getopt.awk BEGIN { while (getopt(ARGC, ARGV, "am:q") != -1) ... } EOF igawk -f test.awk SEE ALSO
gawk(1) Effective AWK Programming, Edition 1.0, published by the Free Software Foundation, 1995. AUTHOR
Arnold Robbins (arnold@skeeve.com). Free Software Foundation Nov 3 1999 IGAWK(1)
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