Grep regex NOT containing string
grep
matches, grep -v
does the inverse. If you need to "match A but not B" you usually use pipes:
grep "${PATT}" file | grep -v "${NOTPATT}"
(?<!1\.2\.3\.4).*Has exploded
You need to run this with -P to have negative lookbehind (Perl regular expression), so the command is:
grep -P '(?<!1\.2\.3\.4).*Has exploded' test.log
Try this. It uses negative lookbehind to ignore the line if it is preceeded by 1.2.3.4
. Hope that helps!
patterns[1]="1\.2\.3\.4.*Has exploded"
patterns[2]="5\.6\.7\.8.*Has died"
patterns[3]="\!9\.10\.11\.12.*Has exploded"
for i in {1..3}
do
grep "${patterns[$i]}" logfile.log
done
should be the the same as
egrep "(1\.2\.3\.4.*Has exploded|5\.6\.7\.8.*Has died)" logfile.log | egrep -v "9\.10\.11\.12.*Has exploded"