Alex Rivera | Logout

Grep regex NOT containing a string

Asked 2012-05-02T10:06:39.923
327

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\.2\.3\.4.*Has exploded" syslog.log

It's just a list of patterns like the "1\.2\.3\.4.*Has exploded" part I am passing, in a loop, so I can't pass "-v", for example.

I am confused trying to do the inverse of the above, and not match lines with a certain IP address and error so "!1.2.3.4.*Has exploded" will match syslog lines for anything other than 1.2.3.4 telling me it has exploded. I must be able to include an IP address to not match.

I have seen various similar posts on Stack Overflow. However, they use regex patterns that I can't seem to get to work with grep. What would be a working example for grep?

This is happening in a script like this;

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
Edit
Report

1 Answer

21
(?<!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 preceded by 1.2.3.4.

answered 2012-05-02T10:12:18.077

Your Answer