logging wrong passwords in /var/log/auth.log












1














/var/log/auth.log logs (among other things) failed login attempts to my debian linux.



I was wandering if it is possible to ask it to log the password that was used in the failed attempt.



This is out of curiosity as to the nature of those failed attempts,
are they using dictionaries? combinations of words? length of passwords used?
and so on



I feel that knowing the passwords that are being tried might help me to better understand the level of risk posed by those attempts.










share|improve this question














bumped to the homepage by Community 17 mins ago


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  • 2




    I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
    – Erathiel
    Apr 30 '15 at 8:21










  • @epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
    – 0xC0000022L
    Apr 30 '15 at 8:25






  • 3




    @0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
    – Romeo Ninov
    Apr 30 '15 at 8:43






  • 1




    @RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
    – 0xC0000022L
    Apr 30 '15 at 10:15






  • 1




    @0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
    – Romeo Ninov
    Apr 30 '15 at 10:20
















1














/var/log/auth.log logs (among other things) failed login attempts to my debian linux.



I was wandering if it is possible to ask it to log the password that was used in the failed attempt.



This is out of curiosity as to the nature of those failed attempts,
are they using dictionaries? combinations of words? length of passwords used?
and so on



I feel that knowing the passwords that are being tried might help me to better understand the level of risk posed by those attempts.










share|improve this question














bumped to the homepage by Community 17 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.











  • 2




    I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
    – Erathiel
    Apr 30 '15 at 8:21










  • @epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
    – 0xC0000022L
    Apr 30 '15 at 8:25






  • 3




    @0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
    – Romeo Ninov
    Apr 30 '15 at 8:43






  • 1




    @RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
    – 0xC0000022L
    Apr 30 '15 at 10:15






  • 1




    @0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
    – Romeo Ninov
    Apr 30 '15 at 10:20














1












1








1







/var/log/auth.log logs (among other things) failed login attempts to my debian linux.



I was wandering if it is possible to ask it to log the password that was used in the failed attempt.



This is out of curiosity as to the nature of those failed attempts,
are they using dictionaries? combinations of words? length of passwords used?
and so on



I feel that knowing the passwords that are being tried might help me to better understand the level of risk posed by those attempts.










share|improve this question













/var/log/auth.log logs (among other things) failed login attempts to my debian linux.



I was wandering if it is possible to ask it to log the password that was used in the failed attempt.



This is out of curiosity as to the nature of those failed attempts,
are they using dictionaries? combinations of words? length of passwords used?
and so on



I feel that knowing the passwords that are being tried might help me to better understand the level of risk posed by those attempts.







security logs password






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Apr 30 '15 at 8:16









epelegepeleg

1064




1064





bumped to the homepage by Community 17 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.







bumped to the homepage by Community 17 mins ago


This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.










  • 2




    I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
    – Erathiel
    Apr 30 '15 at 8:21










  • @epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
    – 0xC0000022L
    Apr 30 '15 at 8:25






  • 3




    @0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
    – Romeo Ninov
    Apr 30 '15 at 8:43






  • 1




    @RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
    – 0xC0000022L
    Apr 30 '15 at 10:15






  • 1




    @0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
    – Romeo Ninov
    Apr 30 '15 at 10:20














  • 2




    I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
    – Erathiel
    Apr 30 '15 at 8:21










  • @epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
    – 0xC0000022L
    Apr 30 '15 at 8:25






  • 3




    @0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
    – Romeo Ninov
    Apr 30 '15 at 8:43






  • 1




    @RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
    – 0xC0000022L
    Apr 30 '15 at 10:15






  • 1




    @0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
    – Romeo Ninov
    Apr 30 '15 at 10:20








2




2




I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
– Erathiel
Apr 30 '15 at 8:21




I'm not sure if this is possible but I'm sure it is definitely not something that even should be possible as this leaks any passwords that are entered with as little mistake as a single typo. If you've ever made a typo while entering your password (and there's no single man on Earth that hasn't) then you'd never even want that to happen.
– Erathiel
Apr 30 '15 at 8:21












@epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
– 0xC0000022L
Apr 30 '15 at 8:25




@epeleg: assuming Linux and other supported platforms, you could presumably write your own PAM module to achieve what you want. Clearly modules like pam_unix.so and pam_cracklib.so have access to the password.
– 0xC0000022L
Apr 30 '15 at 8:25




3




3




@0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
– Romeo Ninov
Apr 30 '15 at 8:43




@0xC0000022L, as fas as I remember they have access to hash, not the password itself. epeleg, from security point of view this is very bad idea.
– Romeo Ninov
Apr 30 '15 at 8:43




1




1




@RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
– 0xC0000022L
Apr 30 '15 at 10:15




@RomeoNinov: I guess your memory betrays you. How would pam_cracklib be able to check for palindromes, passwords containing the username and similar conditions merely based on the hash?
– 0xC0000022L
Apr 30 '15 at 10:15




1




1




@0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
– Romeo Ninov
Apr 30 '15 at 10:20




@0xC0000022L, probably :) But my humble opinion is not to store passwords in log file :)
– Romeo Ninov
Apr 30 '15 at 10:20










1 Answer
1






active

oldest

votes


















0














You can do it the trick mentioned in this blog post,

You edit a line in one of the files of OpenSSH, then compile it and use it.
Short version, just run following script:



OPENSSH=/opt/openssh2
mkdir -p /opt/openssh2/dist/
cd ${OPENSSH}
wget http://zlib.net/zlib-1.2.11.tar.gz
tar xvfz zlib-1.2.11.tar.gz
cd zlib-1.2.11
./configure --prefix=${OPENSSH}/dist/ && make && make install
cd ${OPENSSH}
wget http://www.openssl.org/source/openssl-1.0.1e.tar.gz
tar xvfz openssl-1.0.1e.tar.gz
cd openssl-1.0.1e
./config --prefix=${OPENSSH}/dist/ && make && make install
cd ${OPENSSH}
wget https://ftp.eu.openbsd.org/pub/OpenBSD/OpenSSH/portable/openssh-6.2p1.tar.gz
tar xvfz openssh-6.2p1.tar.gz
cd openssh-6.2p1
sed -e 's/struct passwd * pw = authctxt->pw;/logit("Honey: Username: %s Password: %s", authctxt->user, password);nstruct passwd * pw = authctxt->pw;/' -i auth-passwd.c
./configure --prefix=${OPENSSH}/dist/ --with-zlib=${OPENSSH}/dist --with-ssl-dir=${OPENSSH}/dist/ && make && make install


For the long and interesting version, read the blog post.






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    active

    oldest

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    0














    You can do it the trick mentioned in this blog post,

    You edit a line in one of the files of OpenSSH, then compile it and use it.
    Short version, just run following script:



    OPENSSH=/opt/openssh2
    mkdir -p /opt/openssh2/dist/
    cd ${OPENSSH}
    wget http://zlib.net/zlib-1.2.11.tar.gz
    tar xvfz zlib-1.2.11.tar.gz
    cd zlib-1.2.11
    ./configure --prefix=${OPENSSH}/dist/ && make && make install
    cd ${OPENSSH}
    wget http://www.openssl.org/source/openssl-1.0.1e.tar.gz
    tar xvfz openssl-1.0.1e.tar.gz
    cd openssl-1.0.1e
    ./config --prefix=${OPENSSH}/dist/ && make && make install
    cd ${OPENSSH}
    wget https://ftp.eu.openbsd.org/pub/OpenBSD/OpenSSH/portable/openssh-6.2p1.tar.gz
    tar xvfz openssh-6.2p1.tar.gz
    cd openssh-6.2p1
    sed -e 's/struct passwd * pw = authctxt->pw;/logit("Honey: Username: %s Password: %s", authctxt->user, password);nstruct passwd * pw = authctxt->pw;/' -i auth-passwd.c
    ./configure --prefix=${OPENSSH}/dist/ --with-zlib=${OPENSSH}/dist --with-ssl-dir=${OPENSSH}/dist/ && make && make install


    For the long and interesting version, read the blog post.






    share|improve this answer


























      0














      You can do it the trick mentioned in this blog post,

      You edit a line in one of the files of OpenSSH, then compile it and use it.
      Short version, just run following script:



      OPENSSH=/opt/openssh2
      mkdir -p /opt/openssh2/dist/
      cd ${OPENSSH}
      wget http://zlib.net/zlib-1.2.11.tar.gz
      tar xvfz zlib-1.2.11.tar.gz
      cd zlib-1.2.11
      ./configure --prefix=${OPENSSH}/dist/ && make && make install
      cd ${OPENSSH}
      wget http://www.openssl.org/source/openssl-1.0.1e.tar.gz
      tar xvfz openssl-1.0.1e.tar.gz
      cd openssl-1.0.1e
      ./config --prefix=${OPENSSH}/dist/ && make && make install
      cd ${OPENSSH}
      wget https://ftp.eu.openbsd.org/pub/OpenBSD/OpenSSH/portable/openssh-6.2p1.tar.gz
      tar xvfz openssh-6.2p1.tar.gz
      cd openssh-6.2p1
      sed -e 's/struct passwd * pw = authctxt->pw;/logit("Honey: Username: %s Password: %s", authctxt->user, password);nstruct passwd * pw = authctxt->pw;/' -i auth-passwd.c
      ./configure --prefix=${OPENSSH}/dist/ --with-zlib=${OPENSSH}/dist --with-ssl-dir=${OPENSSH}/dist/ && make && make install


      For the long and interesting version, read the blog post.






      share|improve this answer
























        0












        0








        0






        You can do it the trick mentioned in this blog post,

        You edit a line in one of the files of OpenSSH, then compile it and use it.
        Short version, just run following script:



        OPENSSH=/opt/openssh2
        mkdir -p /opt/openssh2/dist/
        cd ${OPENSSH}
        wget http://zlib.net/zlib-1.2.11.tar.gz
        tar xvfz zlib-1.2.11.tar.gz
        cd zlib-1.2.11
        ./configure --prefix=${OPENSSH}/dist/ && make && make install
        cd ${OPENSSH}
        wget http://www.openssl.org/source/openssl-1.0.1e.tar.gz
        tar xvfz openssl-1.0.1e.tar.gz
        cd openssl-1.0.1e
        ./config --prefix=${OPENSSH}/dist/ && make && make install
        cd ${OPENSSH}
        wget https://ftp.eu.openbsd.org/pub/OpenBSD/OpenSSH/portable/openssh-6.2p1.tar.gz
        tar xvfz openssh-6.2p1.tar.gz
        cd openssh-6.2p1
        sed -e 's/struct passwd * pw = authctxt->pw;/logit("Honey: Username: %s Password: %s", authctxt->user, password);nstruct passwd * pw = authctxt->pw;/' -i auth-passwd.c
        ./configure --prefix=${OPENSSH}/dist/ --with-zlib=${OPENSSH}/dist --with-ssl-dir=${OPENSSH}/dist/ && make && make install


        For the long and interesting version, read the blog post.






        share|improve this answer












        You can do it the trick mentioned in this blog post,

        You edit a line in one of the files of OpenSSH, then compile it and use it.
        Short version, just run following script:



        OPENSSH=/opt/openssh2
        mkdir -p /opt/openssh2/dist/
        cd ${OPENSSH}
        wget http://zlib.net/zlib-1.2.11.tar.gz
        tar xvfz zlib-1.2.11.tar.gz
        cd zlib-1.2.11
        ./configure --prefix=${OPENSSH}/dist/ && make && make install
        cd ${OPENSSH}
        wget http://www.openssl.org/source/openssl-1.0.1e.tar.gz
        tar xvfz openssl-1.0.1e.tar.gz
        cd openssl-1.0.1e
        ./config --prefix=${OPENSSH}/dist/ && make && make install
        cd ${OPENSSH}
        wget https://ftp.eu.openbsd.org/pub/OpenBSD/OpenSSH/portable/openssh-6.2p1.tar.gz
        tar xvfz openssh-6.2p1.tar.gz
        cd openssh-6.2p1
        sed -e 's/struct passwd * pw = authctxt->pw;/logit("Honey: Username: %s Password: %s", authctxt->user, password);nstruct passwd * pw = authctxt->pw;/' -i auth-passwd.c
        ./configure --prefix=${OPENSSH}/dist/ --with-zlib=${OPENSSH}/dist --with-ssl-dir=${OPENSSH}/dist/ && make && make install


        For the long and interesting version, read the blog post.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Aug 28 '18 at 20:29









        FazeLFazeL

        1012




        1012






























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