I would like to list all users in Linux without showing systen-user.
How can I make this only the username .
For example cut -d: -f1 /etc/passwd, I can see all users + system users.
This shows all users with uid less than 999:
awk -F':' '$3>999 {print $1 " uid: " $3}' /etc/passwd | column -t | grep -v nobody
EDIT:
With cut showing only human users:
cut -d: -f1,3 /etc/passwd | egrep ':[0-9]{4}$' | cut -d: -f1
You can try this : awk -F: '$6 ~ /\/home/ {print}' /etc/passwd
Related
i am using "getent passwd | cut -d : -f 1 | xargs groups > users.txt" but this lists all users and the groups they are in. In my case there are thousands of users I only need the list of users part of a specific group.
current output is in pic I don't need the members of linux_users group
Each record in /etc/group ends with a list of users that have that group as their supplementary group ( See http://linux.die.net/man/5/group ).
To get all members of your target group, you need to get those plus all users who have the target group as their primary group (in /etc/passwd, identified by your target group's number mentioned in /etc/group).
This script will get you both (you can call it group_ls or something like that):
#!/bin/sh
num_membs="$( getent group "$1" |cut -d: -f3,4 )"
num="${num_membs%:*}"; membs="${num_membs#*:}"
( IFS=,
printf '%s\n' $membs
awk -v num=$num -F: ' $4 == num { print $1 } ' /etc/passwd
) | sort -u
Alternatively, based on your approach
getent passwd | cut -d : -f 1 | xargs groups |
grep ":.* $1 .*" |cut -d : -f1
should do it too.
You can pipe to grep to narrow down the list that you get. Use some like grep groupname.
Grep: http://www.gnu.org/software/grep/manual/grep.html
Then you can use awk to only get part of a line.
Awk: http://tldp.org/LDP/abs/html/awk.html
So in your case you could do: getent passwd | cut -d : -f 1 | xargs groups | grep linux_admins | awk {'print $1'} > users.txt
You can also read the list of users associated with a group (example: users below) into an array in bash with the following:
IFS=$',\n' read -a names < <(grep '^users:' /etc/group | sed 's/^.*://')
Then you can access each user in a simple loop, e.g.
for i in ${names[#]}; do echo "$i"; done
This will only work for shells that support indexed arrays and process substitution, otherwise the pipe gymnastics are required.
I am using RedHat Linux 6.
I need redhat Linux terminal code to list all the users above uid=499?
i already tried "cat /etc/passwd". but it shows all users. how do i filter it?
You can use awk to parse the passwd database for the UIDs you want.
To list all the users for UIDs strictly greater than 499, do this:
awk -F ':' '$3 > 499' /etc/passwd
EDIT: If you only want the usernames, do this:
getent passwd | awk -F: '$3 > 499 {print $1}'
you can use in your terminal to find all user
cut -d: -f1 /etc/passwd
hope this code help you thanx
I've had a lot of issues with hacks and DDOS attacks on a few servers, though this is usually caused by some very simple things. However I've found it invaluable to be able to look through an accounts access logs and list the hit pages in order of lowest to highest using the following through ssh cat example.co.uk | cut -d\" -f2 | awk '{print $1 " " $2}' | cut -d? -f1 | sort | uniq -c | sort -n
However this means I need to run this against every single accounts access log. is there a server wide version or script out there to scan all access logs for activity?
You can use your command in for loop to check all domain access logs file.
for i in `cat /etc/trueuserdomains | awk {'print $1'} | cut -d":" -f1`;do echo "Pages list Of $i" ; cat /usr/local/apache/domlogs/$i* | grep GET | cut -d\" -f2 | awk '{print $1 " " $2}' | cut -d? -f1 | sort | uniq -c | sort -n;done > /root/report.txt
Once it's done, Please check your /root/report.txt file.
I am working with Linux and I am trying to display and count the users on the system. I am currently using who -q, which gives me a count and the users but I am trying not to list one person more than once with it. At the same time I would like the output of users on separate lines as well or tabbed better than it currently is.
The following will show the number of unique users logged in, ignoring the number of times they are each logged in individually:
who | awk '{ print $1; }' | sort -u | awk '{print $1; u++} END{ print "users: " u}'
If the output of who | awk '{ print $1 }' is :
joe
bunty
will
sally
will
bunty
Then the one-liner will output:
bunty
joe
sally
will
users: 4
Previous answers have involved uniq (but this command only removes duplicates if they are storted, which who does not guarantee, hence we use sort -u to achieve the same.
The awk command at the end outputs the results whilst counting the number of unique users and outputtig this value at the end.
I think you want
who | awk '{print $1}' | uniq && who -q | grep "\# " | cut -d' ' -f2
I'm looking for a way to copy all non-system users from one PC to another. I can get the group and passwd files copied over using this
awk -F":" ' $3 > 499 ' etc/passwd >> /etc/passwd
awk -F":" ' $3 > 499 ' etc/group >> /etc/group
But, how would I go about getting the shadow file copied over since it does not store the UID? Assume that there are over 1000 users, so doing a grep with the usernames, such as egrep '(bob|bill|sarah|sal):' etc/shadow >> /etc/shadow generating the usernames from the awk code above, would be a bit inefficient, but a possible option.
awk -F":" ' $3 > 499 {print "^"$1":"} ' /etc/passwd | sudo grep -f - /etc/shadow > shadow.out
Previous answer could produce multiple lines per user if username is part of other usernames
awk -F":" ' $3 > 499 {print $1} ' /etc/passwd | sudo grep -f - /etc/shadow > shadow.out