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

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/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

















  • 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










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














up vote
1
down vote

favorite












/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

















  • 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










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












up vote
1
down vote

favorite









up vote
1
down vote

favorite











/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













/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




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









epeleg

1064




1064







  • 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










  • @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










  • @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












@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

















up vote
0
down vote













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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    up vote
    0
    down vote













    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
























      up vote
      0
      down vote













      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






















        up vote
        0
        down vote










        up vote
        0
        down vote









        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 at 20:29









        FazeL

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