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Searching for Lines Without a Certain String To search for all the lines of a file that do not contain a certain string, use the -v option to grep .
GNU grep supports three regular expression syntaxes, Basic, Extended, and Perl-compatible. In its simplest form, when no regular expression type is given, grep interpret search patterns as basic regular expressions. To interpret the pattern as an extended regular expression, use the -E ( or --extended-regexp ) option.
To display only the lines that do not match a search pattern, use the -v ( or --invert-match ) option. The -w option tells grep to return only those lines where the specified string is a whole word (enclosed by non-word characters). By default, grep is case-sensitive.
Grep is an implementation of POSIX regular expressions. There are two types of posix regular expressions -- basic regular expressions and extended regular expressions. In grep, generally you use the -E option to allow extended regular expressions.
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"
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