Bash idle session times greater than 15 min - bash

Question

I need help completing this. Trying to take user sessions sitting idle for greater than 15 minutes which aren't being kicked off by sshd_config and kill them. this is what I have to pull the sessions, how do I filter for greater than 15 minutes.
#!/bin/bash
IFS=$'\n'
for output in $(w | tr -s " " | cut -d" " -f1,5 | tail -n+3 | awk '{print $2}')
do
echo "$output \> 15:00"
done

Answer 1

If you are using Awk anyway, a shell loop is a clumsy antipattern. Awk already knows how to loop over lines; use it.
A serious complication is that the output from w is system-dependent and typically reformatted for human legibility.
tripleee$ w | head -n 4
8:16 up 37 days, 19:02, 17 users, load averages: 3.49 3.21 3.11
USER TTY FROM LOGIN# IDLE WHAT
tripleee console - 27Aug18 38days -
tripleee s003 - 27Aug18 38 ssh -t there screen -D -r
If yours looks similar, probably filter out anything where the IDLE field contains non-numeric information
w -h | awk '$5 ~ /[^0-9]/ || $5 > 15'
This prints the entire w output line. You might want to extract just the TTY field ({print $2} on my system) and figure out from there which session to kill.
A more fruitful approach on Linux-like systems is probably to examine the /proc filesystem.

Answer 2

You can try something like this …
for i in $(w --no-header | awk '{print $4}')
do
echo $i | grep days > /dev/null 2>&1
if [ $? == 0 ]
then
echo "greater that 15 mins"
fi
echo $i | grep min> /dev/null 2>&1
if [ $? == 0 ]
then
mins=$(echo $i | sed -e's/mins//g')
if [ $min -gt 15 ]
then
echo "Greater than 15 mins"
fi
fi
done
The tricky part is going to be figuring out what pid to kill.