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Sum out of grep -c

Tags:

linux

I am trying to find the number an even occured in my log file.

Command:

grep -Eo "2016-08-30" applciationLog.log* -c 

Output:

 applciationLog.log.1:0
 applciationLog.log.2:0
 applciationLog.log.3:0
 applciationLog.log.4:0
 applciationLog.log.5:7684
 applciationLog.log.6:9142
 applciationLog.log.7:8699
 applciationLog.log.8:0

What I actually need is sum of all these values 7684 + 9142 + 8699 = 25525. Any suggestion I can do it? Anything I can append to the grep to enable it.

Any help or pointers are welcome and appreciated.

like image 702
user2866507 Avatar asked Dec 01 '25 17:12

user2866507


1 Answers

As addition to the already given answer by ghoti:

You can avoid awk -F: by using grep -h:

grep -c -h -F "2016-08-30" applicationLog.log* | awk '{n+=$0} END {print n}'

This means no filenames and only the counts are printed by grep and we can use the first field for the addition in awk.

like image 57
stackunderglow Avatar answered Dec 04 '25 07:12

stackunderglow



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