[geeks] hrm, possible?

[Bill Bradford]

Given an output like this:

bash-2.02# ps -fe |grep ovw |grep display
 mbarnes 20452     1  0 11:05:34 ?        0:09 /opt/OV/bin/ovw -rw -display 172.17.8.151:0
dtrujill 20078     1  0 09:33:27 ?        0:17 /opt/OV/bin/ovw -ro -display 172.17.11.241:0
lwilliam 20473     1  0 11:10:53 ?        0:10 /opt/OV/bin/ovw -ro -display 172.17.9.166:0
ojanuari 20432     1  0 11:00:44 ?        0:10 /opt/OV/bin/ovw -ro -display 172.17.11.135:0
 dher405 20049     1  0 09:29:07 ?        0:11 /opt/OV/bin/ovw/ -ro -display 172.17.11.136:0
  jwells 20139     1  0 09:47:17 ?        0:19 /opt/OV/bin/ovw -ro -display 172.17.8.206:0
 mbarnes 20029     1 45 09:28:10 ?       11:31 /opt/OV/bin/ovw -rw -display 172.17.8.151:0
 kfuk855 20368     1  0 10:45:05 ?        0:08 /opt/OV/bin/ovw -rw -display 172.17.10.135:0
  proman 20509     1  1 11:31:38 ?        0:09 /opt/OV/bin/ovw -ro -display 172.17.11.64:0

Anybody know if it would be possible to write a script (in whatever
language necessary) that would detect the duplicate sessions (note
processes 20432 and 20368, pointing to the same IP address)?  I'm
working on some way to detect when one of our lusers forgets to 
kill his session before the next shift starts....

Bill (ugh! og programming rusty! og need sharp knife!)

-- 
Bill Bradford
mrbill at mrbill.net
Austin, TX