I have been trying to work out the syntax for this command:
grep ! error_log | find /home/foo/public_html/ -mmin -60
OR:
grep '[^error_log]' | find /home/baumerf/public_html/ -mmin -60
I need to see all files that have been modified except for those named error_log.
I've read about it here, but only found one not-regex pattern.
grep -v is your friend:
grep --help | grep invert
-v, --invert-match select non-matching lines
Also check out the related -L (the complement of -l).
-L, --files-without-match only print FILE names containing no match
You can also use awk for these purposes, since it allows you to perform more complex checks in a clearer way:
Lines not containing foo:
awk '!/foo/'
Lines containing neither foo nor bar:
awk '!/foo/ && !/bar/'
Lines containing neither foo nor bar which contain either foo2 or bar2:
awk '!/foo/ && !/bar/ && (/foo2/ || /bar2/)'
And so on.
grep. Why is this upvoted?
In your case, you presumably don't want to use grep, but add instead a negative clause to the find command, e.g.
find /home/baumerf/public_html/ -mmin -60 -not -name error_log
If you want to include wildcards in the name, you'll have to escape them, e.g. to exclude files with suffix .log:
find /home/baumerf/public_html/ -mmin -60 -not -name \*.log
mmin to search for files modified within 60 mins, use -type f too as mentioned here stackoverflow.com/a/33410471/2361131
Follow WeChat
Success story sharing
-eoption can be used:grep -v -e 'negphrase1' -e 'negphrase2'grep -v 'negphrase1|negphrase2|negphrase3'-Eworks, i.e.grep -vE 'negphrase1|negphrase2|negphrase3'grep "" /dev/null * | grep foo | grep -v bar | cut -d: -f1 | sort -u(why the first grep?, there's always a way :))