Parsing the id command

Shell Programming and Scripting
1

I'm looking to parse the output of the id command.

uid=205(oracle) gid=203(dba) groups=4(adm),207(oinstall),202(ndm),206(eis)

Is there an easy way I can get the user name (in this case "oracle) using a sed or awk command. The username will always be inside the parenthsis
and proceeded with uid=<some number>

Any help would be greatly appreciated.

Thanks in advance to all who answer

2

You could try something along the following lines (I don't know how portable it is, you should check the man pages on your system):

id -un
3

Can't you use the $LOGNAME shell build-in variable ?

echo $LOGNAME

4

It's not portable amongst different flavors of UNIX that's why I dont
want to use it. Thanks for responding.

Ie: on Solaris:

id -un
id: illegal option -- u

5

You could try the POSIX version:

/usr/xpg4/bin/id -un

$ uname -sr
SunOS 5.8
$ id -un
id: illegal option -- u
Usage: id [-ap] [user]
$ PATH="$(getconf PATH)" id -un
drado
6

The username must be the first parenthetical:

awk -F'[()]' '{print $2}'

A smarter approach:

sed 's/^.*uid=[0-9]*(\([^)]*\).*/\1/'

Regards,
Alister

7

Hi,

Another 'sed' solution:

id | sed 's/[^(]\+(\([^)]\+\)).*/\1/'

Regards,
Birei

8

To be specific, a gnu sed solution. \+ is an extension to the posix sed standard.

Regards,
Alister