24 comment

1. I would change it to `cat /etc/passwd | grep -v nologin` which gives a clearer view into which users can actually login and execute.

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   1. Thanks for the feedback! The faq has been updated with more info.

      Reply

   2. This is one of the most common mistakes of unexperimented people on UNIX-like utilities. Instead of ‘cat $file | grep $pattern’ you should use ‘grep $pattern $file’ which is much clearer and allows for easier sudoing.

      Nevertheless, I’ll take note of the ‘nologin’ advice. Thanks!

      Reply

      1. “Instead of ‘cat $file | grep $pattern’ you should use ‘grep $pattern $file’ ”
         I’m one more mistaken user :)

         Reply

2. I would also have a look at:

   last
   lastb
   lastlog

   quite interesting commands on users activity on a host.

   Claudio

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   1. Good call. Appreciate your comment.

      Reply

3. ldap enviroment, whats about:
   # getent passwd |egrep -v ‘nologin|false’
   i recommend this to get a userlist

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4. The ldap and nis scenarios for centralized login administration are omitted here.

   The local passwd file may be just the tip of the login iceberg if either of those is set in
   /etc/nsswitch.conf

   For example:
   # ypcat passwd
   would be the common command if nis is set up.

   Reply

5. new version, with columns

   #!/bin/bash
   # Name: listusers.bash
   # Purpose: List all normal user and system accounts in the system. Tested on RHEL / Debian Linux
   # Author: Vivek Gite , under GPL v2.0+
   # http://www.cyberciti.biz/faq/linux-list-users-command/
   # ———————————————————————————–
   _l=”/etc/login.defs”
   _p=”/etc/passwd”

   ## get mini UID limit ##
   l=$(grep “^UID_MIN” $_l)

   ## get max UID limit ##
   l1=$(grep “^UID_MAX” $_l)

   ## use awk to print if UID >= $MIN and UID = min && $3 = min && $3 <= max && $7 != "/sbin/nologin" ) printf "%-15s %-5s %-5s %-25s %-10sn", $1, $3, $4, $6, $7 }' "$_p"

   echo " "

   ## use awk to print if UID $MAX ##
   echo “—————————-[ System User Accounts ]—————————”
   printf “%-15s %-5s %-5s %-25s %-10sn” “Login” “UID” “GID” “Home” “Shell”
   echo “——————————————————————————-”
   #awk -F’:’ -v “min=${l##UID_MIN}” -v “max=${l1##UID_MAX}” ‘{ if ( !( $3 >= min && $3 = min && $3 <= max && $7 != "/sbin/nologin" ) ) printf "%-15s %-5s %-5s %-25s %-10sn", $1, $3, $4, $6, $7 }' "$_p"

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6. and a group list script

   #!/bin/bash
   # Name: listgroups.bash
   # Purpose: List all normal user and system groups in the system. Tested on RHEL / Debian Linux
   # Author: Vivek Gite , under GPL v2.0+
   # http://www.cyberciti.biz/faq/linux-list-users-command/
   # ———————————————————————————–
   _l=”/etc/login.defs”
   _g=”/etc/group”

   ## get mini GID limit ##
   l=$(grep “^GID_MIN” $_l)

   ## get max GID limit ##
   l1=$(grep “^GID_MAX” $_l)

   ## use awk to print if GID >= $MIN and GID = min && $3 = min && $3 <= max ) printf "%-15s %-5s %-10sn", $1, $3, $4 }' "$_g"

   echo " "

   ## use awk to print if GID $MAX ##
   echo “—————————–[ System User Groups ]—————————-”
   printf “%-15s %-5s %-10sn” “Group” “GID” “Logins”
   echo “——————————————————————————-”
   #awk -F’:’ -v “min=${l##GID_MIN}” -v “max=${l1##GID_MAX}” ‘{ if ( !( $3 >= min && $3 = min && $3 <= max ) ) printf "%-15s %-5s %-10sn", $1, $3, $4 }' "$_g"

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7. humm looks like it got cut.

   drop me an email and ill send both scripts

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8. hi there,

   great script, really helps me a lot, and it’s also very well documented! great work!

   however, since i’m a newbie when it comes to shell scripting, how do I direct the output from stdout into a file?
   I’ve found something like this:
   2>&1 | tee -a users.txt

   but somehow this doesn’t quite work :(

   any help would be appreciated!

   thanks!

   Reply

   1. Redirect the output to a file with the first command and then append the second command output:

      …
      ## use awk to print if UID >= $MIN and UID /tmp/userlist
      awk -F’:’ -v “min=${l##UID_MIN}” -v “max=${l1##UID_MAX}” ‘{ if ( $3 >= min && $3 > /tmp/userlist

      echo “” >> /tmp/userlist
      echo “———-[ System User Accounts ]—————” >> /tmp/userlist
      awk -F’:’ -v “min=${l##UID_MIN}” -v “max=${l1##UID_MAX}” ‘{ if ( !($3 >= min && $3 > /tmp/userlist

      Reply

9. can some body help me with the following question?

   Find the number of users on your system whose user ids are greater than 8?

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10. Thanks. Short and to the point.

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11. Can even use this!
    cat /etc/passwd | grep “/home/” | awk -F’:’ ‘{ print $1}’

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    1. Thanks for this!

       Reply

12. how can see only system users in linux using command

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13. Note that none of this accounts for systems using an external source for its users. If you are using ldap or (gasp!) Active Directory to source your users, then listing /etc/passwd will not yield the desired results, as you won’t see the bulk of your users.

    The command we use, insted of “cat /etc/passwd”, is “getent passwd”, which returns the combined list of users from /etc/passwd (local users) and other sources. The getent command will give you a more realistic view of your users, on any system you encounter.

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    1. Could you show how I would use your command in the final script as posted by the original poster? I think that is the issue I’m having where it is only showing the users with Local Authentication, not AD Users (yes ack.. AD).

       Thanks,
       Bob

       Reply

14. Excellent explanation and script of list users in linux! Congrats Mate!
    Aaron

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15. Thanks for detailed tutorial.

    I configured the VNC as per given step. How to access the same GUI from host machine as we accessing from VNC client.

    Reply

16. Regular user accounts:
    cat /etc/passwd | grep ":[0-9][0-9][0-9][0-9]:"

    System user accounts:
    1) without “nologin”
    cat /etc/passwd | egrep ":[0-9][0-9][0-9]:|:[0-9][0-9]:|:[0-9]:" | egrep - v 'nologin|false'
    2) with “nologin”
    cat /etc/passwd | egrep ":[0-9][0-9][0-9]:|:[0-9][0-9]:|:[0-9]:"

    OR (if you like a separate, clean UIDs listing of one, two or three digits):

    cat /etc/passwd | grep ":[0-9]:"
cat /etc/passwd | grep ":[0-9][0-9]:"
cat /etc/passwd | grep ":[0-9][0-9][0-9]:"

    Reply

17. I actually improved this a little bit; here are some useful aliases which you can put in the ~/.bashrc file:
    # ‘getus’ > Get REGULAR Users
    # ‘getsysus’ > Get SYSTEM Users
    # ‘getgr’ > Get USER Groups
    # ‘getsysgr’ > Get SYSTEM Groups
    # ‘getwheel’ > Get WHEEL’s Users
    alias getus=’getent passwd | egrep “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]”‘
    alias getsysus=’getent passwd | egrep -v “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]”‘
    alias getgr=’getent group | egrep “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]”‘
    alias getsysgr=’getent group | egrep -v “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]”‘
    alias getwheel=’getent group | grep ^wheel’

    Also if you like to ‘egrep’ the ‘nologin’ out of ‘getsysus’ and ‘getsysgr’ above, here you go:
    getsysus | egrep -v nologin
    getsysgr | egrep -v nologin

    Or, simply add them altogether on the above aliases like so:
    alias getsysus=’getent passwd | egrep -v “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]|nologin”‘
    alias getsysgr=’getent group | egrep -v “:[0-9][0-9][0-9][0-9]:|[0-9][0-9][0-9][0-9][0-9]|nologin”‘

    Reply