Issue
My task is to write a bash script, using awk, to find the longest logon of a given user ("still logged in" does not count), and print the month day IP logon time in minutes.
Sample input:./scriptname.sh username1
The content of last username1:
username1 pts/ IP Apr 2 .. .. .. .. (00.03)
username1 pts/ IP Apr 3 .. .. .. .. (00.13)
username1 pts/ IP Apr 5 .. .. .. .. (12.00)
username1 pts/ IP Apr 9 .. .. .. .. (12.11)
Sample output:
Apr 9 IP 731
(expl: 12 hours and 11 minutes is in total 731 minutes)
I have written this script, but a bunch of errors pop up, and I am really confused:
#!/bin/bash
usr=$1
last $usr | grep -v "still logged in" | awk 'BEGIN {max=-1;}
{
h=substr($10,2,2);
min=substr($10,5,2) + h/60;
}
(max < min){
max = min;
}
END{
maxh=max/60;
maxmin=max-maxh;
($maxh == 0 && $maxmin >=10){
last $usr | grep "00:$maxmin" | awk '{print $5," ",$6," ", $3," ",$maxmin}'
exit 1
}
($maxh == 0 $$ $maxmin < 10){
last $usr | grep "00:0$maxmin" | awk '{print $5," ",$6," ",$3," ",$maxmin}'
exit 1
}
($maxh < 10 && $maxmin == 0){
last $usr | grep "0$maxh:00" | awk '{print $5," ",$6," ",$3," ",$maxmin}'
exit 1
}
($maxh < 10 && $maxmin < 10){
last $usr | grep "0$maxh:0$maxmin" | awk '{print $5," ",$6," ",$3," ",$maxmin}'
exit 1
}
($maxh >= 10 && $maxmin < 10){
last $usr | grep "$maxh:0$maxmin" | awk '{print $5," ",$6," ",$3," ",$maxmin}'
exit 1
}
($maxh >=10 && $maxmin >= 10){
last $usr | grep "$maxh:$maxmin" | awk '{print $5," ",$6," ",$3," ",$maxmin}'
exit 1
}
}'
So a bit of explaining of how I imagined this would work:
After the initialization, I want to find the (hh:mm) column of the last $usr command, save the h and min of every line, find the biggest number (in minutes, meaning it is the longest logon time).
After I found the longest logon time (in minutes, stored in the variable max), I then have to reformat the only minutes format to hh:mm to be able to use a grep, use the last command again, but now only searching for the line(s) that contain the max logon time, and print all of the needed information in the month day IP logon time in minutes format, using another awk.
Errors I get when running this code: A bunch of syntax errors when I try using grep and awk inside the original awk.
Solution
awk is not shell. You can't directly call tools like last, grep and awk from awk any more than you could call them directly from a C program.
Using any awk in any shell on every Unix box and assuming if multiple rows have the max time you'd want all of them printed and that if no timestamped rows are found you want something like No matching records printed (easy tweak if not, just tell us your requirements for those cases and include them in the example in your question):
last username1 |
awk '
/still logged in/ {
next
}
{
split($NF,t,/[().]/)
cur = (t[2] * 60) + t[3]
}
cur >= max {
out = ( cur > max ? "" : out ORS ) $4 OFS $5 OFS $3 OFS cur
max = cur
}
END {
print (out ? out : "No matching records")
}
'
Apr 9 IP 731
Answered By - Ed Morton Answer Checked By - Senaida (WPSolving Volunteer)
