用于打印在一个月内的同一时间段内登录系统的平均用户数的脚本

Question

I need to write a script, which prints out the average number of users logged into the system in the same time period in a month. I need to use 'last' and 'awk'.

To make it more clear, let's take for example March and 10:00-11:00 as a time period. Let's suppose on 1st of March we had 2 users logged in that period, on the 2nd of March we had 4 users logged in that period and so on. For those 2 days we have as an average number.

This is what i got so far, but i only managed to display the users logged in a date given.

if [ $# = 0 ]
then
 echo "Give more parameters"
else
 echo "Date:"
 d=`read`
 for i
 do
  echo "user: $i"
  last $i | grep $d
 done
fi

I am expecting to see as a result something like this:

In March the average number of users between 10:00 and 11:00 was x. (just as an example of output expected)

Answer 1

Something of this sort :

last | awk '{
if(NF>7){
if($(NF-5)=="Mar"){ # You may change the month or even accept it as a parameter
if($(NF-4) != date){datecount++};
match($(NF-3),/^([0-9]+)/,arr);
if(arr[1]>=8 && arr[1]<=16){
# Note I am using different time interval, change it accordingly
count++;
}
}
date=$(NF-4);
}
}
END{
print "Total logins   : ",count;
print "Total dates    : ",datecount;
print "Average logins : ",count/datecount;
}'

Sample Output

count     :  20
datecount :  12
average   :  1.66667