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
Related
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
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 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.
I have a number of log files in a directory. I am trying to write a script to search all the log files for a string and echo the name of the files and the line number that the string is found.
I figure I will probably have to use 2 grep's - piping the output of one into the other since the -l option only returns the name of the file and nothing about the line numbers. Any insight in how I can successfully achieve this would be much appreciated.
Many thanks,
Alex
$ grep -Hn root /etc/passwd
/etc/passwd:1:root:x:0:0:root:/root:/bin/bash
combining -H and -n does what you expect.
If you want to echo the required informations without the string :
$ grep -Hn root /etc/passwd | cut -d: -f1,2
/etc/passwd:1
or with awk :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd
file=/etc/passwd
line=1
if you want to create shell variables :
$ awk -F: '/root/{print "file=" ARGV[1] "\nline=" NR}' /etc/passwd | bash
$ echo $line
1
$ echo $file
/etc/passwd
Use -H. If you are using a grep that does not have -H, specify two filenames. For example:
grep -n pattern file /dev/null
My version of grep kept returning text from the matching line, which I wasn't sure if you were after... You can also pipe the output to an awk command to have it ONLY print the file name and line number
grep -Hn "text" . | awk -F: '{print $1 ":" $2}'
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