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to find the number of users logged in

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# 8  
Old 05-30-2007
number of users that logged before me
Hi,

can anybody say --- the no. of users logged before me.
# 9  
Old 05-30-2007
Quote:
Originally Posted by aigles
You have the -x option on (prints executed commands and their arguments).
To disable this option :
Code:
set +x

An example :
Code:
$ set -x
$ date
+ date
Wed May 30 09:40:54 DFT 2007
$ set +x
$ date
Wed May 30 09:41:01 DFT 2007
$

Jean-Pierre
"i want no. of users logged in before me"
# 10  
Old 05-30-2007
Your are continuously bumping up your posts. This is against the rules. Please refrain from breaking the rules.

(4) Do not 'bump up' questions if they are not answered promptly. No duplicate or cross-posting and do not report a post or send a private message where your goal is to get an answer more quickly.
# 11  
Old 05-30-2007
Quote:
Originally Posted by shanshine
"i want no. of users logged in before me"
On my AIX box, the result of who is sorted on the loogin date/time.
Code:
$ who
Name         Line          Time              Hostname
usr001      pts/0       May 24 15:03    (1.2.1.39)  
usr099      pts/1       May 30 07:54    (1.2.1.11)  
usr000      pts/3       May 30 08:58    (1.3.1.2)   
usr099      pts/4       May 30 14:35    (1.2.1.11)  
$

The current terminal is given by the tty command
Code:
$ tty
/dev/pts/4
$

In the output from who, all the users logged in before me are listed before the line with current user and current terminal.
Code:
$ tty=$(tty | sed 's_^/dev/__')
$ echo $tty
pts/4
$ echo $USER
usr099
$ who | awk -v user=$USER -v tty=$tty '$1==user && $2==tty {exit} 1'
usr001      pts/0       May 24 15:03    (1.2.1.39)  
usr099      pts/1       May 30 07:54    (1.2.1.11)  
usr000      pts/3       May 30 08:58    (1.3.1.2)   
$

The final script :
Code:
tty=$(tty | sed 's_^/dev/__')
who | \
awk -v user=$USER -v tty=$tty '$1==user && $2==tty {exit} {print $1}' | \
sort -ud

Jean-Pierre.
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