Linux – Grep last match from file

awkgreplinuxredhat

I'm running the following command against a log file and only wanting to get the last/latest match. Sometimes there may only be one match, other times there may be multiple, which is causing an issue for me since the following command is returning both matches:

cat "$(ls -t | head -n1)" | grep -P "(NODE1[\s\S]*TEST\s=\sPOWER[\s\S]*OUTPUT\s=\s\d+?.*\s+;?)"

>>>>> (results in)...

NODE1 2018-03-06 12:01:23
  TEST = POWER
  EVENT_TIME = 2018-03-06 12:01:23
  OUTPUT = 12

;

NODE1 2018-03-06 12:03:23
  TEST = POWER
  EVENT_TIME = 2018-03-06 12:03:23
  OUTPUT = 7

;

I need the last matching group in the event there are multiple. Is this possible with grep/regex or do I need to pipe the results into sed/awk? If so, how?

Best Answer

I propose to you this solution:

cat <your_source_file> | sed -n '/NODE1/,/;/p' | tr '\n' '|' | awk -F ';' '{print $(NF-1)}'|tr '|' '\n'

sed -n '/NODE1/,/;/p' - find 'NODE1' blocks.

tr '\n' '|' convert newlines to record separator, so table columns will be separated by ';'.

awk -F ';' '{print $(NF-1)}' - print last-1 table column.

tr '|' '\n' - backward to previous view for record.

awk -F ';' '{for(i=(NF-1); i>0; i--){ if($i ~ "TEST = POWER"){print $i} } }' - only "TEST = POWER" events.

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