total login time of each user in last

vril · 04-19-2012, 02:31 PM

Hi everyone. I would like to write a script in bash that displays the total login time of every single user shown by unix "last" command, regardless of any different machines he connects from. How can I achieve that? Thank you.

chrism01 · 04-19-2012, 06:55 PM

Pick your language eg shell, Perl, awk etc, then decide how far back you want to go, then do a trial run with just a few records for test purposes.
Note you'll want to skip no-user entries eg reboot.
What are you going to do if a user has been deleted from the system; will you still count those recs? If not, x-ref against passwd file; might be a good way to skip non-user recs anyway.

Have a go and come back with specific qns and show us your code when you do.

Personally I'd take a copy of the last cmd output rather than pipe it direct into the prog, but that's your call.

Tinkster · 04-19-2012, 08:21 PM

Something like this, maybe? At the mention perl I couldn't resist ;}

Code:

#!/usr/bin/perl 
#===============================================================================
#
#         FILE:  last.pl
#
#        USAGE:  last | ./last.pl  
#
#  DESCRIPTION:  Getting the accumulated login times of users
#===============================================================================

use strict;
use warnings;


my %hours;
my %minutes;
my %logins;
my $user;

while (<STDIN>) {   # While we have input...   
  # Find lines and save username, login time 
  if ( /^(\S+).*\((.*)\)/ && !/^reboot/ ) {
    my $hours;
    my $minutes;
    # jump through a hoop if the user was on for longer than 24 hours
    # in one go ... :)
    my @tmp=split(/:|\+/, $2 );
    # Increment total hours, minutes, and logins 
    if( $#tmp < 2){
      # no overnighters 
      $hours{$1} += $tmp[0];
      $minutes{$1} += $tmp[1]
    }else{
      # overnighters
      $hours{$1} += $tmp[0]*24 +$tmp[1];
      $minutes{$1} += $tmp[2]
    }
    $logins{$1}++;
  }
}

# For each user in the array...        
foreach $user (sort(keys %hours)) {
   # Calculate hours from total minutes 
   $hours{$user} += int($minutes{$user} / 60);
   $minutes{$user} %= 60;
   # Print the information for this user 
   print "User $user, total login time ";
   printf "%02d:%02d, ", $hours{$user}, $minutes{$user};
   print "total logins $logins{$user}.\n";
}

vril · 04-20-2012, 07:31 AM

Guys, sorry but it has to be in bash with use of "last" unix command. Thanks.

Tinkster · 04-20-2012, 04:06 PM

Well, take the logic of what perl does, and replicate it in bash.
Should be possible, maybe slightly more verbose.

danliston · 04-22-2012, 03:19 PM

#!/usr/bin/bash

# Grab a point in time snapshot of the output from the last command
last > /tmp/last_output

# Alternatively because there are output variances to deal with,
last | grep '+' > /tmp/day_more && last | grep -v '+' > /tmp/day_less

# Now we have complete output of last, and split versions of sessions
# that are shorter or longer than one day. We should convert days in
# the day_more file to hours and append to day_less for consistency.
# placeholder to code this thought process...

# we are not interested in 00 minute entries (unless we also decide to count logins)
# We are not interested in any users that are currently logged in or the wtmp footer
# We are not interested in system entries either. Let's hope no users match our egrep
egrep -v '(00:00|logged|^$|wtmp|system)' /tmp/last_output | sed 's/[()]//g' > /tmp/last_clean
cut -d' ' -f1 /tmp/last_clean | sort -u > /tmp/unique_users

# We can crunch data (output) in several ways; Read a line, test each field, process...
# Or we can break our data into separate outputs (by user) before crunching the data.
for user in $(cat /tmp/unique_users)
do
# Guarantee we are starting each loop with clean variables
minute=0
tm=0
hour=0
th=0
day=0
td=0

for time in $(awk /^${user}/'{print $NF}' /tmp/last_clean)
do
temp=$(echo ${time} | cut -d+ -f1)
if [ ${time} = ${temp} ]; then # The day indicator (+) does not exist
day=0
hour=$(echo ${time} | cut -d: -f1 | sed 's/^0//')
else
day=$(echo ${time} | cut -d+ -f1)
hour=$(echo ${time} | cut -d+ -f2 | cut -d: -f1 | sed 's/^0//')
fi
# We use sed to strip padded 0's in single digit values for math (07 becomes just 7)
minute=$(echo ${time} | cut -d: -f2 | sed 's/^0//')
((td = td + day))
((th = th + hour))
((tm = tm + minute))
done
# We have our column totals, but now we need to normalize the output again
minute=$(expr $tm % 60)
hour=$(expr `echo "$tm / 60 + $th" | bc` % 24)
day=$(expr `echo "$tm / 60 + $th" | bc` / 24 + $td)
printf "%8s %4d + %02d : %02d \n" $user $day $hour $minute
done

# Do I get an A?