How to exclude words using grep?

code-8

11/13/23, 5:00 PM

Instead of me chaining these like 10 times

cat /var/log/nginx/access.log | grep -v download | grep -v danger | grep -v "/visitor/return" | grep -v welcome | grep -v "GET / HTTP" | grep -v "/baby/" | grep -v "/paste/" | awk '{ print $1, $4, $7 }' >> access.$(date "+%Y-%m-%d").txt

Is there a way to combine them into a list or an array for a bit cleaner to read ?

I tried

grep -wv 'download|danger|/visitor/return|welcome|GET / HTTP|/baby/|/paste/' | awk '{ print $1, $4, $7 }' >> access.$(date "+%Y-%m-%d").txt

no luck, even with -v only

I thinking I'm setting here for the best

cat /var/log/nginx/access.log | grep -v -e "download" -e "danger" -e "/visitor/return" -e "welcome" -e "GET / HTTP" -e "/baby/" -e "/paste/" | awk '{ print $1, $4, $7 }' >> access.$(date "+%Y-%m-%d").txt

If you guys know a better one, I would like to learn from you. !

366

1 + 0

command-line

bash

grep

Score:0

Ubuntu

kanehekili

11/13/23, 7:26 PM

You may use the -Ev flag:

grep -Ev 'house|garden' file.txt

will show all lines that do not the words "house" or "garden".

-E: Interpret PATTERN as an extended regular expression
-v: Invert the sense of matching, to select non-matching lines.

An alternative would be the -v flag with egrep:

egrep -v '(house|garden)' file.txt

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