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

8
By: eanx
2009-03-03 04:26:57

What Others Think

Try replacing less with wc -l to get the number of invalid users.
aperson · 506 weeks and 4 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
atoponce · 506 weeks and 3 days ago
somehow missed the '.0' on the second auth.log. you get the idea.
atoponce · 506 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
raphink · 506 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.)
OJM · 506 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
dbart · 506 weeks and 3 days ago

What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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