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Parsing the id command

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# 1  
Old 10-12-2010
Parsing the id command
I'm looking to parse the output of the id command.

Code:
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
Last edited by radoulov; 10-12-2010 at 03:56 PM.. Reason: Added code tags!
# 2  
Old 10-12-2010
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):

Code:
id -un

# 3  
Old 10-12-2010
Can't you use the $LOGNAME shell build-in variable ?

echo $LOGNAME
Last edited by ctsgnb; 10-12-2010 at 04:29 PM..
# 4  
Old 10-12-2010
Quote:
Originally Posted by radoulov
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):

Code:
id -un

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  
Old 10-12-2010
You could try the POSIX version:

Code:
/usr/xpg4/bin/id -un



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

# 6  
Old 10-12-2010
The username must be the first parenthetical:
Code:
awk -F'[()]' '{print $2}'

A smarter approach:
Code:
sed 's/^.*uid=[0-9]*(\([^)]*\).*/\1/'

Regards,
Alister
# 7  
Old 10-13-2010
Hi,

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

Regards,
Birei
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