How to retrieve IP addresses of possible ssh attackers?











up vote
17
down vote

favorite
10












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









share|improve this question




























    up vote
    17
    down vote

    favorite
    10












    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









    share|improve this question


























      up vote
      17
      down vote

      favorite
      10









      up vote
      17
      down vote

      favorite
      10






      10





      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









      share|improve this question















      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






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      share|improve this question













      share|improve this question




      share|improve this question








      edited Mar 18 '15 at 9:00

























      asked Mar 18 '15 at 8:22









      kravemir

      1,42621938




      1,42621938






















          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.






          share|improve this answer





















          • 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








          • 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 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




            What about IPv6?
            – Ortomala Lokni
            Apr 14 '16 at 16:31


















          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






          share|improve this answer





















          • 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


















          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





          share|improve this answer




























            up vote
            0
            down vote













            grep "Failed password for" /var/log/auth.log |
            awk -F"from" {'print $2'} |
            awk {'print $1'} |
            sort -u





            share|improve this answer























            • 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










            • 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













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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.






            share|improve this answer





















            • 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








            • 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 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




              What about IPv6?
              – Ortomala Lokni
              Apr 14 '16 at 16:31















            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.






            share|improve this answer





















            • 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








            • 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 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




              What about IPv6?
              – Ortomala Lokni
              Apr 14 '16 at 16:31













            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.






            share|improve this answer












            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.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            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 -n to 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 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




              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










            • 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




              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 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




              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












            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






            share|improve this answer





















            • 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















            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






            share|improve this answer





















            • 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













            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






            share|improve this answer












            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







            share|improve this answer












            share|improve this answer



            share|improve this answer










            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


















            • 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










            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





            share|improve this answer

























              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





              share|improve this answer























                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





                share|improve this answer












                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






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Mar 19 '15 at 0:08









                Falsenames

                63748




                63748






















                    up vote
                    0
                    down vote













                    grep "Failed password for" /var/log/auth.log |
                    awk -F"from" {'print $2'} |
                    awk {'print $1'} |
                    sort -u





                    share|improve this answer























                    • 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










                    • 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

















                    up vote
                    0
                    down vote













                    grep "Failed password for" /var/log/auth.log |
                    awk -F"from" {'print $2'} |
                    awk {'print $1'} |
                    sort -u





                    share|improve this answer























                    • 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










                    • 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















                    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





                    share|improve this answer














                    grep "Failed password for" /var/log/auth.log |
                    awk -F"from" {'print $2'} |
                    awk {'print $1'} |
                    sort -u






                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Dec 2 at 23:30









                    roaima

                    42.5k551116




                    42.5k551116










                    answered Mar 18 '15 at 8:54









                    Liviu Popescu

                    112




                    112












                    • 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










                    • 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




















                    • 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










                    • 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


















                    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




















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