please create a file called “ myloginfile ” as shown below and then write an awk

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 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

Explanation / Answer

Idea:

First subset the time based on mins.. Then, check if the hours are zero. If hours are zero then it is less then one hour. Hence, print that line.

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.