Reputation: 447
I have a data file something like following. I want to grep "PANIC:", "Node 0/1/2/3:" and blank lines, but want to include next line ( "-A1" option ) only to one pattern - "PANIC:", not others. How to do that.
Kindly help to find solution
# cat data.txt
Node 0:
DATE: 2020-12-19 04:06:44
DATE: 2020-12-19 07:02:54
PANIC: ABC
DATE: 2020-12-19 20:02:54
DATE: 2020-12-18 21:25:39
Node 1:
DATE: 2020-12-20 09:35:43
DATE: 2020-12-20 12:12:29
Node 2:
Node 3:
DATE: 2020-12-19 20:03:00
DATE: 2020-12-19 15:35:09
PANIC: XYZ
DATE: 2020-12-19 06:44:27
#
# cat data.txt | grep -A1 --no-group-separator -e PANIC: -e 'Node [0-7]:' -e ^$
Node 0:
DATE: 2020-12-19 04:06:44
PANIC: ABC
DATE: 2020-12-19 20:02:54
Node 1:
DATE: 2020-12-20 09:35:43
Node 2:
DATE: 2020-12-20 03:00:15
Node 3:
DATE: 2020-12-19 20:03:00
PANIC: XYZ
DATE: 2020-12-19 06:44:27
#
Expected filtered data is something like this
Node 0:
PANIC: ABC
DATE: 2020-12-19 20:02:54
Node 1:
Node 2:
Node 3:
PANIC: XYZ
DATE: 2020-12-19 06:44:27
Upvotes: 1
Views: 177
Reputation: 212198
You can do as you ask with:
awk '/PANIC:/{print; getline; print} /Node [0-7]/ || /^ *$/' input
Although you don't mention it in the question, your sample output demonstrates that you also want to combine multiple blank lines in a single blank line. That can be done with:
awk '/./{a=0} /PANIC:/{print; getline; print} /^ *$/ && !a++{print} /Node [0-7]/' input
To print the line before the PANIC match, it's probably easiest to simply record the previous line. Something like:
awk '/PANIC:/{print prev; print; getline; print} { prev = $0 } /Node [0-7]/ || /^ *$/' input | perl -00pe1
Upvotes: 1
Reputation: 8711
Using awk
awk -F: ' /Node|PANIC/ { print; if($1~/PANIC/) { getline; print} } ' data.txt
Results:
$ awk -F: ' /Node|PANIC/ { print; if($1~/PANIC/) { getline; print} } ' susanta.txt
Node 0:
PANIC: ABC
DATE: 2020-12-19 20:02:54
Node 1:
Node 2:
Node 3:
PANIC: XYZ
DATE: 2020-12-19 06:44:27
$
Upvotes: 1