How to retrieve IP addresses of possible ssh attackers?
up vote
17
down vote
favorite
I've just cat /var/log/auth.log log and see, that there are many | grep "Failed password for" records.
However, there are two possible record types - for valid / invalid user. It complicates my attempts to | cut them.
I would like to see create a list (text file) with IP addresses of possible attackers and number of attempts for each IP address. Is there any easy way to create it?
Also, regarding only ssh: What all records of /var/log/auth.log should I consider when making list of possible attackers?
Example of my 'auth.log' with hidden numbers:
cat /var/log/auth.log | grep "Failed password for" | sed 's/[0-9]/1/g' | sort -u | tail
Result:
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user ucpss from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user vijay from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user webalizer from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user xapolicymgr from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user yarn from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zookeeper from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zt from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for mysql from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.111.111.1 port 11111 ssh1
ssh scripting security
add a comment |
up vote
17
down vote
favorite
I've just cat /var/log/auth.log log and see, that there are many | grep "Failed password for" records.
However, there are two possible record types - for valid / invalid user. It complicates my attempts to | cut them.
I would like to see create a list (text file) with IP addresses of possible attackers and number of attempts for each IP address. Is there any easy way to create it?
Also, regarding only ssh: What all records of /var/log/auth.log should I consider when making list of possible attackers?
Example of my 'auth.log' with hidden numbers:
cat /var/log/auth.log | grep "Failed password for" | sed 's/[0-9]/1/g' | sort -u | tail
Result:
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user ucpss from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user vijay from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user webalizer from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user xapolicymgr from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user yarn from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zookeeper from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zt from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for mysql from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.111.111.1 port 11111 ssh1
ssh scripting security
add a comment |
up vote
17
down vote
favorite
up vote
17
down vote
favorite
I've just cat /var/log/auth.log log and see, that there are many | grep "Failed password for" records.
However, there are two possible record types - for valid / invalid user. It complicates my attempts to | cut them.
I would like to see create a list (text file) with IP addresses of possible attackers and number of attempts for each IP address. Is there any easy way to create it?
Also, regarding only ssh: What all records of /var/log/auth.log should I consider when making list of possible attackers?
Example of my 'auth.log' with hidden numbers:
cat /var/log/auth.log | grep "Failed password for" | sed 's/[0-9]/1/g' | sort -u | tail
Result:
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user ucpss from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user vijay from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user webalizer from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user xapolicymgr from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user yarn from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zookeeper from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zt from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for mysql from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.111.111.1 port 11111 ssh1
ssh scripting security
I've just cat /var/log/auth.log log and see, that there are many | grep "Failed password for" records.
However, there are two possible record types - for valid / invalid user. It complicates my attempts to | cut them.
I would like to see create a list (text file) with IP addresses of possible attackers and number of attempts for each IP address. Is there any easy way to create it?
Also, regarding only ssh: What all records of /var/log/auth.log should I consider when making list of possible attackers?
Example of my 'auth.log' with hidden numbers:
cat /var/log/auth.log | grep "Failed password for" | sed 's/[0-9]/1/g' | sort -u | tail
Result:
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user ucpss from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user vijay from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user webalizer from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user xapolicymgr from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user yarn from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zookeeper from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for invalid user zt from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for mysql from 111.111.11.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.11.111.111 port 11111 ssh1
Mar 11 11:11:11 vm11111 sshd[111]: Failed password for root from 111.111.111.1 port 11111 ssh1
ssh scripting security
ssh scripting security
edited Mar 18 '15 at 9:00
asked Mar 18 '15 at 8:22
kravemir
1,42621938
1,42621938
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
up vote
19
down vote
accepted
You could use something like this:
grep "Failed password for" /var/log/auth.log | grep -Po "[0-9]+.[0-9]+.[0-9]+.[0-9]+"
| sort | uniq -c
It greps for the string Failed password for and extracts (-o) the ip address. It is sorted, and uniq counts the number of occurences.
The output would then look like this (with your example as input file):
1 111.111.111.1
3 111.11.111.111
6 111.111.11.111
The last one in the output has tried 6 times.
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added| sort -nto the chain.
– kravemir
Mar 18 '15 at 10:42
1
Good answer - multiple greps are usually a sign to use sed.sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}'replaces both greps.
– orion
Mar 18 '15 at 10:57
1
@orion true but imagine that the first grep can also bezgrep "Failed" /var/log/auth.log*to also search trough the compressed log archives, whatsedcannot.
– chaos
Mar 18 '15 at 11:02
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
add a comment |
up vote
6
down vote
It might be a bloated solution but i suggest you look at installing something like Fail2Ban
It is made for this kind of logging + adds the bonus of being able to add (temporary) rules in your firewall to block repeating offenders. Be sure to whitelist your own ip's though, i managed to lock myself out temporarily on a few occasions
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
add a comment |
up vote
0
down vote
This worked out real well for me. (IPs have been changed to protect the guilty)
$ awk '/Failed/ {x[$(NF-3)]++} END {for (i in x){printf "%3d %sn", x[i], i}}' /var/log/auth.log | sort -nr
65 10.0.0.1
14 10.0.0.2
4 10.0.0.3
1 10.0.0.4
add a comment |
up vote
0
down vote
grep "Failed password for" /var/log/auth.log |
awk -F"from" {'print $2'} |
awk {'print $1'} |
sort -u
The result ispassword- not working. It doesn't even cover different entry types,awkprint column is alternative to thecut, not the solution.
– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it missesuniq -cas in the best answer.
– kravemir
Mar 18 '15 at 10:39
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
19
down vote
accepted
You could use something like this:
grep "Failed password for" /var/log/auth.log | grep -Po "[0-9]+.[0-9]+.[0-9]+.[0-9]+"
| sort | uniq -c
It greps for the string Failed password for and extracts (-o) the ip address. It is sorted, and uniq counts the number of occurences.
The output would then look like this (with your example as input file):
1 111.111.111.1
3 111.11.111.111
6 111.111.11.111
The last one in the output has tried 6 times.
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added| sort -nto the chain.
– kravemir
Mar 18 '15 at 10:42
1
Good answer - multiple greps are usually a sign to use sed.sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}'replaces both greps.
– orion
Mar 18 '15 at 10:57
1
@orion true but imagine that the first grep can also bezgrep "Failed" /var/log/auth.log*to also search trough the compressed log archives, whatsedcannot.
– chaos
Mar 18 '15 at 11:02
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
add a comment |
up vote
19
down vote
accepted
You could use something like this:
grep "Failed password for" /var/log/auth.log | grep -Po "[0-9]+.[0-9]+.[0-9]+.[0-9]+"
| sort | uniq -c
It greps for the string Failed password for and extracts (-o) the ip address. It is sorted, and uniq counts the number of occurences.
The output would then look like this (with your example as input file):
1 111.111.111.1
3 111.11.111.111
6 111.111.11.111
The last one in the output has tried 6 times.
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added| sort -nto the chain.
– kravemir
Mar 18 '15 at 10:42
1
Good answer - multiple greps are usually a sign to use sed.sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}'replaces both greps.
– orion
Mar 18 '15 at 10:57
1
@orion true but imagine that the first grep can also bezgrep "Failed" /var/log/auth.log*to also search trough the compressed log archives, whatsedcannot.
– chaos
Mar 18 '15 at 11:02
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
add a comment |
up vote
19
down vote
accepted
up vote
19
down vote
accepted
You could use something like this:
grep "Failed password for" /var/log/auth.log | grep -Po "[0-9]+.[0-9]+.[0-9]+.[0-9]+"
| sort | uniq -c
It greps for the string Failed password for and extracts (-o) the ip address. It is sorted, and uniq counts the number of occurences.
The output would then look like this (with your example as input file):
1 111.111.111.1
3 111.11.111.111
6 111.111.11.111
The last one in the output has tried 6 times.
You could use something like this:
grep "Failed password for" /var/log/auth.log | grep -Po "[0-9]+.[0-9]+.[0-9]+.[0-9]+"
| sort | uniq -c
It greps for the string Failed password for and extracts (-o) the ip address. It is sorted, and uniq counts the number of occurences.
The output would then look like this (with your example as input file):
1 111.111.111.1
3 111.11.111.111
6 111.111.11.111
The last one in the output has tried 6 times.
answered Mar 18 '15 at 9:10
chaos
35k773115
35k773115
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added| sort -nto the chain.
– kravemir
Mar 18 '15 at 10:42
1
Good answer - multiple greps are usually a sign to use sed.sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}'replaces both greps.
– orion
Mar 18 '15 at 10:57
1
@orion true but imagine that the first grep can also bezgrep "Failed" /var/log/auth.log*to also search trough the compressed log archives, whatsedcannot.
– chaos
Mar 18 '15 at 11:02
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
add a comment |
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added| sort -nto the chain.
– kravemir
Mar 18 '15 at 10:42
1
Good answer - multiple greps are usually a sign to use sed.sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}'replaces both greps.
– orion
Mar 18 '15 at 10:57
1
@orion true but imagine that the first grep can also bezgrep "Failed" /var/log/auth.log*to also search trough the compressed log archives, whatsedcannot.
– chaos
Mar 18 '15 at 11:02
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
this is the best answer @chaos - added to my box of useful oneliners - thanks!
– Jake
Mar 18 '15 at 9:56
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added
| sort -n to the chain.– kravemir
Mar 18 '15 at 10:42
Great solution. I didn't know that grep can extract regex matches, not just filter lines. I've just added
| sort -n to the chain.– kravemir
Mar 18 '15 at 10:42
1
1
Good answer - multiple greps are usually a sign to use sed.
sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}' replaces both greps.– orion
Mar 18 '15 at 10:57
Good answer - multiple greps are usually a sign to use sed.
sed -nr '/Failed/{s/.*([0-9]+.[0-9]+.[0-9]+.[0-9]+).*/1/;p}' replaces both greps.– orion
Mar 18 '15 at 10:57
1
1
@orion true but imagine that the first grep can also be
zgrep "Failed" /var/log/auth.log* to also search trough the compressed log archives, what sed cannot.– chaos
Mar 18 '15 at 11:02
@orion true but imagine that the first grep can also be
zgrep "Failed" /var/log/auth.log* to also search trough the compressed log archives, what sed cannot.– chaos
Mar 18 '15 at 11:02
1
1
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
What about IPv6?
– Ortomala Lokni
Apr 14 '16 at 16:31
add a comment |
up vote
6
down vote
It might be a bloated solution but i suggest you look at installing something like Fail2Ban
It is made for this kind of logging + adds the bonus of being able to add (temporary) rules in your firewall to block repeating offenders. Be sure to whitelist your own ip's though, i managed to lock myself out temporarily on a few occasions
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
add a comment |
up vote
6
down vote
It might be a bloated solution but i suggest you look at installing something like Fail2Ban
It is made for this kind of logging + adds the bonus of being able to add (temporary) rules in your firewall to block repeating offenders. Be sure to whitelist your own ip's though, i managed to lock myself out temporarily on a few occasions
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
add a comment |
up vote
6
down vote
up vote
6
down vote
It might be a bloated solution but i suggest you look at installing something like Fail2Ban
It is made for this kind of logging + adds the bonus of being able to add (temporary) rules in your firewall to block repeating offenders. Be sure to whitelist your own ip's though, i managed to lock myself out temporarily on a few occasions
It might be a bloated solution but i suggest you look at installing something like Fail2Ban
It is made for this kind of logging + adds the bonus of being able to add (temporary) rules in your firewall to block repeating offenders. Be sure to whitelist your own ip's though, i managed to lock myself out temporarily on a few occasions
answered Mar 18 '15 at 9:07
Jake
36839
36839
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
add a comment |
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
Well, nice commentary, but not a answer to the question. It's more like suggestion that could remove the need of question, but I don't need a daemon to monitor my files. I've got my reasons why do I need to make that list as text file just by scripts. :)
– kravemir
Mar 18 '15 at 13:11
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
I agree with @Miro, had you not mentioned it here I'd have added it in a comment.
– SailorCire
Mar 18 '15 at 15:35
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
@Miro, you're right, not an answer to the question itself, it just came to mind as a convenient tool for this kind of logging.
– Jake
Mar 19 '15 at 15:27
add a comment |
up vote
0
down vote
This worked out real well for me. (IPs have been changed to protect the guilty)
$ awk '/Failed/ {x[$(NF-3)]++} END {for (i in x){printf "%3d %sn", x[i], i}}' /var/log/auth.log | sort -nr
65 10.0.0.1
14 10.0.0.2
4 10.0.0.3
1 10.0.0.4
add a comment |
up vote
0
down vote
This worked out real well for me. (IPs have been changed to protect the guilty)
$ awk '/Failed/ {x[$(NF-3)]++} END {for (i in x){printf "%3d %sn", x[i], i}}' /var/log/auth.log | sort -nr
65 10.0.0.1
14 10.0.0.2
4 10.0.0.3
1 10.0.0.4
add a comment |
up vote
0
down vote
up vote
0
down vote
This worked out real well for me. (IPs have been changed to protect the guilty)
$ awk '/Failed/ {x[$(NF-3)]++} END {for (i in x){printf "%3d %sn", x[i], i}}' /var/log/auth.log | sort -nr
65 10.0.0.1
14 10.0.0.2
4 10.0.0.3
1 10.0.0.4
This worked out real well for me. (IPs have been changed to protect the guilty)
$ awk '/Failed/ {x[$(NF-3)]++} END {for (i in x){printf "%3d %sn", x[i], i}}' /var/log/auth.log | sort -nr
65 10.0.0.1
14 10.0.0.2
4 10.0.0.3
1 10.0.0.4
answered Mar 19 '15 at 0:08
Falsenames
63748
63748
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add a comment |
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0
down vote
grep "Failed password for" /var/log/auth.log |
awk -F"from" {'print $2'} |
awk {'print $1'} |
sort -u
The result ispassword- not working. It doesn't even cover different entry types,awkprint column is alternative to thecut, not the solution.
– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it missesuniq -cas in the best answer.
– kravemir
Mar 18 '15 at 10:39
add a comment |
up vote
0
down vote
grep "Failed password for" /var/log/auth.log |
awk -F"from" {'print $2'} |
awk {'print $1'} |
sort -u
The result ispassword- not working. It doesn't even cover different entry types,awkprint column is alternative to thecut, not the solution.
– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it missesuniq -cas in the best answer.
– kravemir
Mar 18 '15 at 10:39
add a comment |
up vote
0
down vote
up vote
0
down vote
grep "Failed password for" /var/log/auth.log |
awk -F"from" {'print $2'} |
awk {'print $1'} |
sort -u
grep "Failed password for" /var/log/auth.log |
awk -F"from" {'print $2'} |
awk {'print $1'} |
sort -u
edited Dec 2 at 23:30
roaima
42.5k551116
42.5k551116
answered Mar 18 '15 at 8:54
Liviu Popescu
112
112
The result ispassword- not working. It doesn't even cover different entry types,awkprint column is alternative to thecut, not the solution.
– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it missesuniq -cas in the best answer.
– kravemir
Mar 18 '15 at 10:39
add a comment |
The result ispassword- not working. It doesn't even cover different entry types,awkprint column is alternative to thecut, not the solution.
– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it missesuniq -cas in the best answer.
– kravemir
Mar 18 '15 at 10:39
The result is
password - not working. It doesn't even cover different entry types, awk print column is alternative to the cut, not the solution.– kravemir
Mar 18 '15 at 8:56
The result is
password - not working. It doesn't even cover different entry types, awk print column is alternative to the cut, not the solution.– kravemir
Mar 18 '15 at 8:56
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
It works for me either with sample above ou real /var/log/auth.log
– Archemar
Mar 18 '15 at 9:31
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it misses
uniq -c as in the best answer.– kravemir
Mar 18 '15 at 10:39
Well, it didn't work before the edit - when the comment had been made. Now it works correctly. However, it misses
uniq -c as in the best answer.– kravemir
Mar 18 '15 at 10:39
add a comment |
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