how to know how many user are using now the connector ?

[Michael REMY]

hi!

we have 100users and 50 licenses for outlook connector.
how to know in real time (command line by example), the number of conncted users who are using the connector now ?

thanks for help

[odeleon]

You could grep mailbox.log and look for the OlkConnector string (I think that's how it's spelled)

[Michael REMY]

thanks for your reply.

i am not an expert, do you have the right command line ?

[Goonsniper]

Hi

I created this to grep the mailbox.log file located in /opt/zimbra/log/ to find which version of the ZCO users are running. This will, of course, only find users which have contacted the server in the current log file.

grep ZCO mailbox.log |awk -F[ '{print $3}' | awk -F= '{print $2 $5}' |awk -F\;mid '{print $1" " $2}' | awk -F\( '{print $1}' |uniq |more

It produces output like...

user1@domain.com Zimbra-ZCO/6.0.5846.5
user2@domain.com Zimbra-ZCO/5.0.3064.18
user3@domain.com Zimbra-ZCO/6.0.5981.7
user4@domain.com Zimbra-ZCO/6.0.5981.7

Hope this helps.

[fyd]

grep ZCO mailbox.log |awk -F[ '{print $3}' | awk -F= '{print $2 $5}' |awk -F\;mid '{print $1" " $2}' | awk -F\( '{print $1}' |uniq |more

It produces output like...

user1@domain.com Zimbra-ZCO/6.0.5846.5
user2@domain.com Zimbra-ZCO/5.0.3064.18
user3@domain.com Zimbra-ZCO/6.0.5981.7
user4@domain.com Zimbra-ZCO/6.0.5981.7

Hope this helps.

Good one goonsniper. Thanks. But I guess the uniq there doesn't effectively filter out duplicate lines of with the account name. Mailbox log has multiple lines that contains the account name while carrying out a particular connector event. I just tried and came across this.

user1@domain.com Zimbra-ZCO/5.0.2992.16
user1@domain.com ZimbraConnectorForOutlook/5.0.2992.16;] calendar - Fixed up ORGANIZER in a REPLY from ZCO
user1@domain.com Zimbra-ZCO/5.0.2992.16

Tried playing some switches with uniq and sort. Didn't work I think you'll know better.

[LMStone]

Some activity not having to do with the ZCO Connector in the mailbox log will contain the characters "ZCO", being able to run this handy script from anywhere would be nice, being able to see who the very active users are, and eliminating the duplicates, which can be done by piping the output through "sort", would also be nice, so I offer the minor modifications below:

grep Zimbra-ZCO /opt/zimbra/log/mailbox.log |awk -F[ '{print $3}' | awk -F= '{print $2 $5}' |awk -F\;mid '{print $1" " $2}' | awk -F\( '{print $1}' |sort |uniq -c |more

Works for us, but please test and post if it works for you!

Hope that helps,
Mark

[fyd]

Hi Mark, yes it works funky. Thanks.
And thanks to goonsniper, yours just needed a minor tuning.

[rayb2001]

Here's a shell script and an awk script that shows the number of users and the ZCO version. Let me know if you find any improvements.

#! /bin/bash

zgrep ZimbraConnectorForOutlook /opt/zimbra/log/mailbox.* | grep "soap - SyncRequest" | awk -f ZCO.awk > /tmp/ZCO_Users
chmod 666 /tmp/ZCO_Users
less /tmp/ZCO_Users

Then, here's the awk script (as referred to above as ZCO.awk):

BEGIN { i=0 }
{
 if ( split ($0, s1, "[") != 3 ) next                   # If the input is malformed, skip this line
    n = split (s1[3], s2, ";")                          # Sometimes the mid= is missing so account for that
    split (s2[1], uname, "=")                           # Get the email addr
    split (s2[n-1], ua, "=")                            # Get the ua= string
    split (ua[2], uav, "/")                             # Separate out only the version
    result[i++] = sprintf ("%s\t%s", uname[2], uav[2])  # Save the result to put in an array
}

END {
  if (i == 0) exit 1

  n = asort(result)     # sort the results

  cnt = 0               # count how many unique users
  for (i = 1; i < n; i++) {
        if (result[i] == result[i+1]) continue          # skip duplicates
        print (result[i])
        cnt++
  }
  print (result[n])     # print the last one (cnt does not include this one)
  printf ("-----------------------------------------------\n")
  printf ("Total ZCO users: %d\n", cnt+1)               # Print the total
  exit 0
}

[btriem]

Thanks for taking the time to present to us this script. It's just extremely sad that it's not built-in somehow already since it is how they define 'professional' users...

[fyd]

Thanks for taking the time to present to us this script. It's just extremely sad that it's not built-in somehow already since it is how they define 'professional' users...

Hi btriem, I think you should go ahead filing an RFE at bugzilla under requests, and get votes for the feature.