Emergency UNIX and Linux Support

As I have shown you, I could see the output from the code...
Still I suggest, please show us the entire output as is of "lsuser <username>", we shall try to help

---------- Post updated at 02:57 PM ---------- Previous update was at 02:54 PM ----------

Try this and confirm

Code:

lsuser <username> | awk '/gecos=/ {print $NF}' FS=';'

or

Code:

lsuser <username> | awk '/^[ \t]*gecos=/ {print $NF}' FS=';'

What you call "gecos" field usually is the "user name or comment field", field 5 in the : separated /etc/passwd file. Try

Code:

awk -F: '{sub(/^.*;/,"",$5);print $5}' /etc/passwd

From the command "lsuser" and the general output format i deduce it is an AIX system (still, it would be nice to state that instead of letting the audience exert its collective criminalistic skills). The output of "lsuser" is a space-separated list of user properties in the form "property=value".

It is possible to limit the output to one (or several) specific properties, therefore:

Code:

lsuser -a gecos <username>

will limit the output to only the gecos property (and the username, which is always part of the output). From there it should be easy to extract everything up to the delimiting semicolon by shell expansion:

Code:

var="$(lsuser -a gecos <username>)"
print - ${var#*;}

I hope this helps.

bakunin

Code:

cat /etc/passwd | awk -F: '($1=="'$1'"){print $5}' | awk -F\; '{print $3}'

That is a useless use of cat. Also, if you're doing awk | awk | awk, you might as well put it all in one awk.

Also, never use creative quoting to get variables into awk, that's unsafe and there's no need.

Code:

awk -F: '($1==U){sub(/.*;/, "", $5); print $5 }' U="$1" /etc/passwd

try this

Code:

lsuser -a gecos <username> | cut -d ';' -f3

gecos, not "user name or comment field" is the correct name for this field. Entries in this field should be separated by commas and such entries can be changed by a user using the chfn(1) utility.