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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 6 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 339 weeks 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 339 weeks ago

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

Comment by atoponce 339 weeks 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 339 weeks 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 339 weeks 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 339 weeks ago

Your point of view

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