Run tail -f for a specific time in bash script

问题

I need a script that will run a series of tail -f commands and output them into a file. What I need is for tail -f to run for a certain amount of time to grep specific words. The reason it's a certain amount of time is because some of these values don't show up right away as this is a live log.

How can I run something like this for let's say 20 seconds, output the grep command and then continue on to the next command?

tail -f /example/logs/auditlog | grep test

Thanks

回答1:

timeout 20 tail -f /example/logs/auditlog | grep test

回答2:

tail -f /example/logs/auditlog | grep test &
pid=$!
sleep 20
kill $pid

回答3:

What about this:

for (( N=0; $N < 20 ; N++)) ; do tail -f /example/logs/auditlog | grep test ; sleep 1 ; done

EDIT: I misread your question, sorry. You want something like this:

tail -f /example/logs/auditlog | grep test
sleep 20

来源：https://stackoverflow.com/questions/16368382/run-tail-f-for-a-specific-time-in-bash-script