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Extract user accounts and home directory from /etc/passwd.


 
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# 1  
Old 06-10-2014
Extract user accounts and home directory from /etc/passwd.

I am trying to obtain all user accounts and their respective home directories.

/etc/passwd contains the required information, but I want to filter it to only show the uid,username and home directory path.

I am working on a Solaris 11 machine.

I made a little headway so far, but I got stuck Smilie.
Code:
cat /etc/passwd | cut -f1,3,6 -d:

I have to eliminate results which have >100 uid, as they are not user accounts.

I tried using grep to do so, but it got really out of hand lol Smilie

Code:
cat /etc/passwd | cut -f1,3,6 -d: | grep '[0-9][0-9][0-9]:[a-z]+:/[a-z]+/[a-z]+/[a-z]+'

Any help is greatly appreciated.
# 2  
Old 06-10-2014
Code:
awk -F: ' $3 <= 100 {print $1, $6}' /etc/passwd

---------- Post updated at 11:16 PM ---------- Previous update was at 11:00 PM ----------

Code:
while IFS=: read user shadow uid guid geco home sh; do if [[ $uid -le 100 ]]; then echo $user $home; fi; done < /etc/passwd

This User Gave Thanks to Aia For This Post:
# 3  
Old 06-10-2014
Thanks a million, your 2nd script worked. I don't know what it is, but it is magical.

I actually wanted >100, so I changed le 100 to ge 100.

Cheers Smilie
# 4  
Old 06-10-2014
Quote:
Originally Posted by Hijanoqu
Thanks a million, your 2nd script worked. I don't know what it is, but it is magical.

I actually wanted >100, so I changed le 100 to ge 100.

Cheers Smilie
>100 is $uid -gt 100; $uid -ge 100 is >= 100.
# 5  
Old 06-10-2014
Quote:
Originally Posted by Hijanoqu
I tried using grep to do so, but it got really out of hand lol Smilie

Code:
cat /etc/passwd | cut -f1,3,6 -d: | grep '[0-9][0-9][0-9]:[a-z]+:/[a-z]+/[a-z]+/[a-z]+'

I think you got offered a better solution here, but in case your curiosity is still aroused, this is how grep/cut can do it:

grep's task is to select all lines where UID is below 100. I have no Solaris system at hand, so here is a passwd from an AIX system. The ideas how to filter are the same, though.

Code:
# cat /etc/passwd
root:!:0:0::/:/usr/bin/ksh
daemon:!:1:1::/etc:
bin:!:2:2::/bin:
sys:!:3:3::/usr/sys:
adm:!:4:4::/var/adm:
uucp:!:5:5::/usr/lib/uucp:
nobody:!:4294967294:4294967294::/:
lpd:!:9:4294967294::/:
lp:*:11:11::/var/spool/lp:/bin/false
invscout:*:6:12::/var/adm/invscout:/usr/bin/ksh
nuucp:*:7:5:uucp login user:/var/spool/uucppublic:/usr/sbin/uucp/uucico
snapp:*:200:13:snapp login user:/usr/sbin/snapp:/usr/sbin/snappd
ipsec:*:201:1::/etc/ipsec:/usr/bin/ksh
sshd:*:202:201::/var/empty:/usr/bin/ksh
opc_op:*:777:202:OVO default operator:/home/opc_op:/usr/bin/sh
apache:*:64500:64500:Apache User:/home/apache:/usr/bin/ksh
wikiroot:*:153:1:Wiki Admin:/var/www/htdocs:/usr/bin/ksh
mysql:*:251:251:MySQL system user:/opt/freeware/mysql:/usr/bin/ksh
ctmagp:*:38713:5003:CM Agent User:/home/ctmagp:/bin/ksh
tsmbkusr:*:38638:1:TSM-Backup User:/home/tsmbkusr:/bin/ksh
plotadm:*:252:1:tech. user fuer plotgenerierung:/home/plotadm:/usr/bin/ksh
nmon2rrd:*:253:1:user fuer generierung von nmon2rrd daten:/home/nmon2rrd:/usr/bin/ksh
collect:*:204:1:data collector:/home/collect:/usr/bin/ksh
lpar2rrd:*:205:1:tech user lpar2rrd:/var/datasource/lpar2rrd:/usr/bin/ksh

First, what is "lower than/greater than 100" in regexps? Below 100:
Code:
/[0-9]*[0-9]/

This covers every number with one or two digits 0-9, therefore "0-99". We can use "grep -v" to invert that regexp and get evrything greater/equal. Because this is the third field and is delimited by ":" we add these to the regexp to further reduce ambiguity:

Code:
/:[0-9]*[0-9]:/

Now, because we want this to only match the third field, not any other, what is the "third field" here? It is: some characters, followed by a colon, followed by some more characters, followed by a colon - and now any character up to the next colon. In regexp:

Code:
/^[^:]*:[^:]*:[0-9]*[0-9]:/

Now try it:

Code:
grep '^[^:]*:[^:]*:[0-9]*[0-9]:' /etc/passwd | cut -d':' -f1,3,6          # below UID 100
grep -v '^[^:]*:[^:]*:[0-9]*[0-9]:' /etc/passwd | cut -d':' -f1,3,6       # UID 100 and above

I hope this helps.

bakunin
These 2 Users Gave Thanks to bakunin For This Post:
# 6  
Old 06-10-2014
I'm afraid that [0-9]*[0-9] will match any integer number, even longer than two digits. Try [0-9]?[0-9] instead. You may need to escape the quotation mark or use extended grep.
This User Gave Thanks to RudiC For This Post:
# 7  
Old 06-10-2014
Quote:
Originally Posted by RudiC
I'm afraid that [0-9]*[0-9] will match any integer number, even longer than two digits.
Yes, you are right. My bad. Here is a corrected version of the sensitive part of the regexp:

Code:
/[0-9]\{1,2\}/

bakunin
 

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