在 bash 脚本中运行 tail -f 特定时间
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Run tail -f for a specific time in bash script
提问by user2348517
I need a script that will run a series of tail -fcommands and output them into a file.
What I need is for tail -fto 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.
我需要一个脚本来运行一系列tail -f命令并将它们输出到一个文件中。我需要的是tail -f运行一定的时间来 grep 特定的单词。之所以有一定的时间，是因为其中一些值不会立即显示，因为这是实时日志。
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?
我怎样才能运行这样的事情，比如说 20 秒，输出 grep 命令，然后继续执行下一个命令？
tail -f /example/logs/auditlog | grep test
Thanks
谢谢
回答by Karoly Horvath
timeout 20 tail -f /example/logs/auditlog | grep test
回答by Chris Dodd
tail -f /example/logs/auditlog | grep test &
pid=$!
sleep 20
kill $pid
回答by hannenz
What about this:
那这个呢：
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:
编辑：我误读了你的问题，抱歉。你想要这样的东西：
tail -f /example/logs/auditlog | grep test
sleep 20
