Question: please create a file called myloginfile as shown below and then write an awk script to determine how many logins in file myloginfile spent less
please create a file called myloginfile as shown below and then write an awk script to determine how many logins in file myloginfile spent less than an hour on the system.
$ last | grep "Oct 20" > myloginfile
cfs264fa pts/3 c-68-46-14-229.h Tue Oct 20 23:24 - 23:47 (00:22)
cfs264fa pts/0 c-68-46-14-229.h Tue Oct 20 22:51 - 23:24 (00:33)
cfs264fa pts/0 c-68-46-14-229.h Tue Oct 20 22:43 - 22:45 (00:02)
ics325su pts/0 75-168-124-222.m Tue Oct 20 21:57 - 22:15 (00:17)
ics325fa pts/0 c-98-240-232-78. Tue Oct 20 16:08 - 20:33 (04:25)
cfs264fa pts/3 173-8-126-169-mi Tue Oct 20 14:57 - 16:07 (01:09)
cfs264fa pts/0 12.28.108.200 Tue Oct 20 11:33 - 15:15 (03:42)
cfs264fa pts/0 c-24-118-47-197. Tue Oct 20 05:49 - 06:14 (00:25)
Please provide your awk script below along with a screenshot of the execution of the script:
This is what i have so far but i am running into a syntax error because of the ':'
awk { if ( $10 < 01:00) print $10} myloginfile
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