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Printing a block of lines from a file, if that block does not contain two patterns using sed

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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
# 1  
Old 12-21-2012
Linux Printing a block of lines from a file, if that block does not contain two patterns using sed
I want to process a file block by block using sed, and if that block does not contain two patterns, then that complete block has to be printed.

See below for the example data.
Code:
................................server 1...............................
running                     process 1
running                     process 2
running                     process 3
date : 12/12/2012
............................................................................
................................server 2...............................
running                     process 1
stopped                    process 2
running                     process 3
date : 13/12/2012
............................................................................

in the above mentioned file a block refers from
.........server................. to ........................
in that block i have to check, if there is no pattern "stop" and "13/12/2012" then i have to print that block.

this is what i have tried till now

Code:
sed -n -e '/./{H;/\.\.\.\./!d;}' -e 'x;/stop/ !{ /13.12.2012/ !{p;}}' example

the output iam getting is given below
Code:
 ................................server 1...............................
................................server 1...............................
running                     process 1
running                     process 2
running                     process 3
date : 12/12/2012
............................................................................
............................................................................
................................server 2...............................

The required output is
Code:
 ................................server 1...............................
running                     process 1
 running                     process 2
 running                     process 3
 date : 12/12/2012
 ............................................................................

Please let me know what am I doing wrong.Smilie
Last edited by Scrutinizer; 12-21-2012 at 09:32 AM.. Reason: icode to code tagss
# 2  
Old 12-21-2012
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

Last edited by Scrutinizer; 12-21-2012 at 10:02 AM..
# 3  
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.
# 4  
Old 12-21-2012
I don't understand, there is no "11.12.2012" in your sample, I get:
Code:
$ awk 'p{p=p RS $0} /\.\.\.\./ {if(p~s && !(p~/stop/)){print p; p=x}else {p=$0}}' s="12.12.2012" file
................................server 1...............................
running                     process 1
running                     process 2
running                     process 3
date : 12/12/2012
$

Code:
$ awk 'p{p=p RS $0} /\.\.\.\./ {if(p~s && !(p~/stop/)){print p; p=x}else {p=$0}}' s="13.12.2012" file
$

# 5  
Old 12-21-2012
Is this what you want to do?

Code:
$ awk -v d=$(date +%Y/%m/%d) '{p=$0~/stopped/?s=1:p=p RS $0}NR%6==0{if(!s)print p;p=s=x}/date/{if($3==d)s=1}' file

................................server 1...............................
running                     process 1
running                     process 2
running                     process 3
date : 12/12/2012
............................................................................

This only works if every block has 6 lines though.
Last edited by Subbeh; 12-21-2012 at 10:44 AM.. Reason: forgot about the date part
# 6  
Old 12-21-2012
Perhaps you mean this:
Code:
$ awk 'p{p=p RS $0} /\.\.\.\./ {if(p){ if(p!~s && p!~/stop/)print p; p=x}else {p=$0}}' s="13.12.2012" file
................................server 1...............................
running                     process 1
running                     process 2
running                     process 3
date : 12/12/2012
............................................................................
$

This User Gave Thanks to Scrutinizer For This Post:
Kesavan
# 7  
Old 12-21-2012
I am not sure if this is correct:
Code:
sed -n '/^\.\.\.\./!H; /^\.\.\.\./{x; /stopped/d; /date.*13\/12\/2012/d; p;}' example

These 2 Users Gave Thanks to MadeInGermany For This Post:
Kesavan sathyaonnuix
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