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need a one liner to grep a group info from /etc/group and use that result to search passwd file

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Top Forums Shell Programming and Scripting need a one liner to grep a group info from /etc/group and use that result to search passwd file
# 1  
Old 02-08-2012
need a one liner to grep a group info from /etc/group and use that result to search passwd file
/etc/group
Code:
tiadm::345:mk789,po312,jo343,ju454,ko453,yx879,iy345,hn453
bin::2:root,daemon
sys::3:root,bin,adm
adm::4:root,daemon
uucp::5:root

/etc/passwd
Code:
mk789:x:234:1::/export/home/dummy:/bin/sh
po312:x:234:1::/export/home/dummy:/bin/sh
ju454:x:234:1::/export/home/dummy:/bin/sh
ko453:x:234:1::/export/home/dummy:/bin/sh
yx879:x:234:1::/export/home/dummy:/bin/sh
iy345:x:234:1::/export/home/dummy:/bin/sh
hn453:x:234:1::/export/home/dummy:/bin/sh
root:x:0:0:Super-User:/:/sbin/sh
daemon:x:1:1::/:
bin:x:2:2::/usr/bin:
sys:x:3:3::/:
adm:x:4:4:Admin:/var/adm:
lp:x:71:8:Line Printer Admin:/usr/spool/lp:
uucp:x:5:5:uucp Admin:/usr/lib/uucp:
nuucp:x:9:9:uucp Admin:/var/spool/uucppublic:/usr/lib/uucp/uucico

Now i need to search for tiadm group and then extract the users in that group and use it to find their info in password file

Please suggest some one liner. I am using solaris box

---------- Post updated at 12:58 AM ---------- Previous update was at 12:56 AM ----------

My try..

grep 'tiadm' /etc/group | awk -F: '{print $4}' | awk F, {print $1} --> after this i am blank

after this i am not sure how to use it.

i can do this with shell script but i am searching for one liner
# 2  
Old 02-08-2012
There could be more efficient ways. But I'm just extending your attempt to a working solution:
Code:
grep `grep 'tiadm' /etc/group  | awk -F: '{print $4}' | awk -F, '{print $1}'` /etc/passwd

Here's another:
Code:
grep `grep 'tiadm' /etc/group  | sed 's/.*:\([^,]*\),.*/\1/'` /etc/passwd

Last edited by balajesuri; 02-08-2012 at 02:09 AM..
This User Gave Thanks to balajesuri For This Post:
chidori
# 3  
Old 02-08-2012
well that will just print password user information of one user.
# 4  
Old 02-08-2012
One awk:
Code:
awk -F: 'NR==FNR && $1==g{split($4,T,","); for(i in T) U[T[i]]; next} $1 in U' g=tiadm /etc/group /etc/passwd

This User Gave Thanks to Scrutinizer For This Post:
chidori
# 5  
Old 02-08-2012
Quote:
Now i need to search for tiadm group and then extract the users in that group and use it to find their info in password file
Wonder what you mean by "their info"?

If it's just the login name:
Code:
logins -g tiadm | awk '{print $1}'

If it's everything available and in a machine-readable format:
Code:
logins -xtog tiadm

This User Gave Thanks to methyl For This Post:
chidori
# 6  
Old 02-08-2012
Thanks scrutinizer code works..

Thanks metyl i didnt know a system command could do the work Smilie
one more query,i have seen in the password logins doesnt give correct information. i used to depend on passwd -s command to know the status.
# 7  
Old 02-08-2012
The "logins" command does give the correct information about password status etc and you get more information than "passwd -s". Just beware of the date format (mmddyy).
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