Get Unique IP Address and # of Access

Question

I found the following code to see which IP Addresses has the highest hits:

FILE=access.log; for ip in cat $FILE |cut -d ' ' -f 1 |sort |uniq; do { COUNT=grep ^$ip $FILE |wc -l; if [[ "$COUNT" -gt "500" ]]; then echo "$COUNT: $ip"; fi }; done

The above code displays the IP address with more than 500 hits (i.e. access on the site by opening the URL)

But that script is too slow. Is there any other code that create the same output?

Plus, how to display the top 10 results only that has the highest hits or access on the URL?

Short answer: yes. `cut -d' ' -f1 access.log | sort | uniq -c | awk '$1 ~ /[5-9][0-9][0-9]|[0-9][0-9][0-9][0-9]+/'`. Only reads through your access log once rather than once per IP address. You can add `| sort -n` on the end if you want them sorted by hits. — Ladadadada, Jul 11 '13 at 15:23

score 1 · Answer 1 · answered Jul 11 '13 at 14:00

1

You're re-inventing the wheel. Try this:

sed -e 's/\([0-9]\+\.[0-9]\+\.[0-9]\+\.[0-9]\+\).*$/\1/' -e t -e d access.log | sort | uniq -c

answered Jul 11 '13 at 14:00

Satanicpuppy

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Thanks. How to get top 10 or top 20 only with highest hits? – jaYPabs Jul 11 '13 at 14:02
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Just add '| head -n 20' to the end of the command above. – slhsen Dec 30 '15 at 13:56

Get Unique IP Address and # of Access

1 Answers1