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Quickly find a count of how many times invalid users have attempted to access your system

Terminal - Quickly find a count of how many times invalid users have attempted to access your system
gunzip -c /var/log/auth.log.*.gz | cat - /var/log/auth.log /var/log/auth.log.0 | grep "Invalid user" | awk '{print $8;}' | sort | uniq -c | less
2009-03-03 04:26:57
User: eanx
Functions: awk cat grep gunzip sort uniq
8
Quickly find a count of how many times invalid users have attempted to access your system

Alternatives

There are 3 alternatives - vote for the best!

Terminal - Alternatives

Know a better way?

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What others think

Try replacing less with wc -l to get the number of invalid users.

Comment by aperson 298 weeks and 3 days ago

Use 'zcat' instead of 'gunzip -c' and there is no use of that 'grep' statement. Awk is grep on steroids, so take advantage of it.

zcat /var/log/auth.log.*.gz | cat - /var/log/auth.log /var/log/auth.log | awk '/Invalid user/ {print $8;}' | sort | uniq -c | less
Comment by atoponce 298 weeks and 3 days ago

somehow missed the '.0' on the second auth.log. you get the idea.

Comment by atoponce 298 weeks and 3 days ago

You could also sort it again in the end :

zcat /var/log/auth.log.*.gz | cat - /var/log/auth.log /var/log/auth.log | awk '/Invalid user/ {print $8;}' | sort | uniq -c | sort -n | less
Comment by raphink 298 weeks and 3 days ago

zgrep "Invalid User" /var/log/auth.log* | awk '{print $8}' | sort | uniq -c | sort -nr

Much faster. (zgrep will happily grep through files which are not compressed.)

Comment by OJM 298 weeks and 3 days ago

OJM, you had a typo (User instead of user), so yours is:

zgrep "Invalid user" /var/log/auth.log* | awk '{print $8}' | sort | uniq -c | sort -nr | less

Also, you might want to look for times when legit users failed to login:

zgrep "Failed password" /var/log/auth.log* | awk '{print $9}' | sort | uniq -c | sort -nr | less

Comment by dbart 298 weeks and 2 days ago

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