Grep regex NOT containing string

Question

I am passing a list of regex patterns to grep to check against a syslog file. They are usually matching an IP address and log entry;

grep "1         


        

Answer 1

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"    

Answer 2

(?<!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!

Answer 3

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}"