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Find who logged in system apart from myself

 
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# 1  
Old 07-04-2011
Find who logged in system apart from myself
Hi,

I need to find the users logged in the system beside me.
as

Code:
uname -u

gives all the user and

Code:
uname -um

gives the current user on system.

How can i get result of
Code:
uname -u

minus
Code:
uname -um

.

I want to do it in one line.

tried with grep but not successful.
# 2  
Old 07-04-2011
Use who and id to stay portable:
Code:
who | grep -v "^$( id -un )"

# 3  
Old 07-04-2011
Thanks pludi.

but the problem is :

like the user "root" is being used by many users to login and i want to find that apart from me(current user ) who else have logged in with root.So just excluding the name of user doesn't help Smilie.
# 4  
Old 07-04-2011
If possible post sample output of uname -u and uname -um and the machine your working.
# 5  
Old 07-04-2011
Here is the output i get:

Code:
who -u
kailash  pts/0        2011-07-04 10:41   .          8336 (172.31.169.1)
kailash  pts/1        2011-07-04 10:41 03:13        8442 (172.31.169.1)
kailash  pts/3        2011-07-04 15:20 00:48       11094 (172.31.169.1)
kailash  pts/4        2011-07-04 15:20 01:35       11197 (172.31.169.1)
kailash  pts/5        2011-06-28 11:30 04:18       13030 (172.31.168.146)

Code:
kailash  pts/0        2011-07-04 10:41   .          8336 (172.31.169.1)


Its a sample output.
# 6  
Old 07-05-2011
Can we grep out the . and try..?
Code:
# Machine - GNU Linux's output
who -a | grep -v " \. "
who -a | awk '$6!="."{print}'

This User Gave Thanks to michaelrozar17 For This Post:
kailash19
# 7  
Old 07-05-2011
Thanks Michael.

It works Smilie Smilie.

but for :
Code:
who -a | awk '$6!="."{print}'

it should be :
Code:
who -a | awk '$5!="."{print}'

Also i have thought of this one:
Code:
WHO=`who -um`;who -u | grep -v "$WHO"

 
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