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
Related
I am using grep to parse the password file.
When I use
grep -w "avahi" /etc/passwd
I get two responses
avahi and avahi-autoipd
I have not found a method to give me the unique response.
This command is part of a bigger script where the name (avahi) is actually a variable.
This does work when the name is rpc and rpcuser. So I am guessing the it has something to do with the dash (-) in the name.
Actual code:
#!/bin/ksh
getent shadow |cut -d: -f1-2|grep ':!!'| cut -d: -f1 > /tmp/pasck
while read line
do
NOLOGIN=`grep -w $line /etc/passwd | cut -d -f7|cut -d/ -f3`
if [[ $NOLOGIN != "nologin: && $NOLOGIN != "false" ]] ; then
echo "$line" "$NOLOGIN" >> /tmp/pasck.list
fi
done <?tmp/pasck
The script is trying to go through the shadow file and look for users with no passwords. Then I compare the results to the passwd file to find which of those accounts are set to /bin/false or /sbin/nologin. The remainder would be actual users with no password set but allowed on the system.
Keeping things simple - you could include colon that comes after the username in your grep statement:
$ grep "^avahi:" /etc/passwd
You can use awk and explicitly test the first colon-separated field:
awk -F: '$1 = "avahi"' /etc/passwd
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
Considering that all users with id >= 1000 are non-system users, how can we get list of these users in a single command?
You need to get all users whose gid is greater than or equals 1000. Use this command for that:
awk -F: '($3>=1000)&&($1!="nobody"){print $1}' /etc/passwd
If you want system users (gid<1000) it will be:
awk -F: '($3<1000){print $1}' /etc/passwd
You can use awk for this task:
awk -F: '$3 >= 1000' /etc/passwd
This will split the /etc/passwd file by colon, then if field 3 (userid) is greater than or equal to 1000, it will print the entire /etc/passwd record.
If you want to get only the username out of this list then:
awk -F: '$3 >= 1000 {print $1}' /etc/passwd
Where $1 is the first field of etc/passwd which is the username.
Supposing that the system recognizes only local users (i.e. those recorded in /etc/passwd, as opposed to any authenticated via a remote service such as LDAP, NIS, or Winbind), you can use grep, sed, or awk to extract the data from /etc/passwd. awk is the most flexible of those, but how about a solution with sed:
sed -n '/^\([^:]\+\):[^:]\+:[1-9][0-9]\{3\}/ { s/:.*//; p }' /etc/passwd
System users (should be) those listed in /etc/passwd with UIDs less than 1000. The actual number is a convention only. Non-system users need not be listed there. You can get the list using getent and awk ignoring "nobody" (also a convention):
getent passwd |awk -F : '$3 >= 1000 && $3 < 65534'
Here's an answer for doing this on all your machines, using Ansible and awk, building on JNevill's answer:
ansible -i inventories/cd_staging all -m shell -a "awk -F: '\$3 >= 1000 && \$7 \!~ /nologin/ {print \$1}' \/etc\/passwd |sort"
You'll want to ignore GIDs less than 1000, but also GIDs greater than 60000. Ubuntu/Debian reserve these for various system services.
awk -F: '($3>=1000)&&($3<60000)&&($1!="nobody"){print $1}' /etc/passwd
I want to display all users that its UID between 300 and 500.
I tried the grep command but I can't get the result that I need.
I tried this syntax, but it does not work:
cat /etc/passwd | grep *:[300-500]
Using awk, here is your answer:
awk -F: '$3 < 500 && $3 > 300 { print $0 }' /etc/passwd
You can print $1 if you just want the username.
egrep 'x:3[0-9][0-9]:|x:4[0-9][0-9]:|x:500:' /etc/passwd
or more eloquent
egrep 'x:[3-4][0-9][0-9]:|x:500:' /etc/passwd
I need a grep or another like command that will pull the usernames only from the /etc/passwd file in linux. Anything before the colon. I know this is doing with reg ex however I am not nearly experienced enough...
The following command will give all ACTUAL users, I need a way to pipe to grep or another line of code to only display the username portion.
awk -v LIMIT=500 -F: '{print $1}' '($3>=LIMIT) && ($3!=65534)' /etc/passwd
This should do:
awk -F: '$3>=LIMIT && $3!=65534 {print $1}' LIMIT=500 /etc/passwd
To do this in mostly plain bash:
limit=500
nfsnobody_id=65534
cut -d: -f1,3 /etc/passwd | while IFS=: read username uid; do
(( uid >= limit && uid != nfsnobody_id )) && echo $username
done
Get out of the habit of using VARNAMES_IN_CAPS: one day you'll write PATH=$(dirname $FILE) and then wonder why commands can no longer be found.
You simply need to rewrite the filter part of your awk command a bit. For instance, the following should work:
awk -v LIMIT=500 -F: '{if (($3>=LIMIT) && ($3!=65534)) print $1}' /etc/passwd
Otherwise, to answer your question strictly, if you want to use grep, the command would be
... | grep -o "^[^:]*"
but that would not be the way to go.