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I'm writing script is ksh. Need to find all users who has over N process and echo them in shell. N reads from ksh.

I know what I should use ps -elf but how parse it, find users with >N process and create array with them. Little troubles with array in ksh. Please help. Maybe simple solutions can help me instead of array creating.

s162103@helios:/home/s162103$ ps -elf 0 S s153308 4804 1 0 40 20 ? 17666 ? 11:03:08 ? 0:00 /usr/lib/gnome-settings-daemon --oa 0 S root 6546 1327 0 40 20 ? 3584 ? 11:14:06 ? 0:00 /usr/dt/bin/dtlogin -daemon -udpPor 0 S webservd 15646 485 0 40 20 ? 2823 ? п╪п╟я─я ? 0:23 /opt/csw/sbin/nginx 0 S s153246 6746 6741 0 40 20 ? 18103 ? 11:14:21 ? 0:00 iiim-panel --disable-crash-dialog 0 S s153246 23512 1 0 40 20 ? 17903 ? 09:34:08 ? 0:00 /usr/bin/metacity --sm-client-id=de 0 S root 933 861 0 40 20 ? 5234 ? 10:26:59 ? 0:00 dtgreet -display :14 ...

when i type 

ps -elf | awk '{a[$3]++;}END{for(i in a)if (a[i]>N)print i, a[i];}' N=1 s162103@helios:/home/s162103$ ps -elf | awk '{a[$3]++;}END{for(i in a)if (a[i]>N)print i, a[i];}' N=1 root 118 /usr/sadm/lib/smc/bin/smcboot 3 /usr/lib/autofs/automountd 2 /opt/SUNWut/lib/utsessiond 2 nasty 31 dima 22 /opt/oracle/product/Oracle_WT1/ohs/ 7 /usr/lib/ssh/sshd 5 /usr/bin/bash 11

that is not user /usr/sadm/lib/smc/bin/smcboot there is last field in ps -elf ,not user

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3 Answers 3

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Something like this(assuming 3rd field of your ps command gives the user id):

 ps -elf | awk '{a[$3]++;} END { for(i in a) if (a[i]>N) print i, a[i]; }' N=3
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yes 3rd field,but it outputs wrong information. It output random information(UID and CMD and other from ps -elf),not only names I need only names.(UIDs)
F S UID PID PPID C PRI NI ADDR SZ WCHAN STIME TTY TIME CMD 5 S cl32387 18246 18241 0 80 0 - 5895 ? Mar16 ? 00:00:00 /usr/sbin/vsftpd /etc/vsftpd/vs5 S cl32387 18469 18436 0 80 0 - 13200 ? 09:51 ? 00:00:00 sshd: cl32387@pts/2 0 S cl32387 18481 18469 0 80 0 - 4470 - 09:51 pts/2 00:00:00 -bash 0 R cl32387 18842 18481 0 80 0 - 3431 - 09:51 pts/2 00:00:00 ps -elf
I am getting proper output for the same..keep in mind, the solution provided will list all the users and their no. of processes if it is more than 3(N=3).
There's my edited post on the top. Guru, please,check it. There is information about output ps -elf ant your command in formatted
with nawk it works fine but number of count is wrong. For example: root 136. ps -elf show less than 136
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The minimal ps command you want to use here is ps -eo user=. This will just print the username for each process and nothing more. The rest can be done with awk:

ps -eo user= | awk -v max=3 '{ n[$1]++ } END { for (user in n) if (n[user]>max) print n[user], user }'

I recommend to put the count in the first column for readability.

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read number ps -elfo user= | sort | uniq -c | while read count user do if (( $count > $number )) then echo $user fi done

That is best solution and it works!

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