如何优雅地停止bash脚本中的无限循环? [英] How to stop infinite loop in bash script gracefully?

问题描述

我需要每隔X秒运行一次应用程序，因此，就cron不能以这种方式运行秒而言，我编写了一个bash脚本，其中包含X睡眠的无限循环.

I need to run application in every X seconds, so, as far as cron does not work with seconds this way, I wrote a bash script with infinite loop having X seconds sleep in it.

当我必须手动停止正在运行的脚本时，我想以一种正确的方式进行操作-让应用程序完成运行，而下次不再进入循环.

When I have to stop the running script manually, I would like to do it in a correct way - let the application complete functioning and just do not enter the loop for the next time.

您有什么想法，如何实现? 我曾考虑过传递参数，但找不到如何将参数传递给正在运行的脚本.

Do you have any idea, how this can be achieved? I thought about passing arguments, but I could not find how to pass argument to running script.

推荐答案

您可以捕获信号，例如SIGUSR1:

You could trap a signal, say SIGUSR1:

echo "My pid is: $$"
finish=0
trap 'finish=1' SIGUSR1

while (( finish != 1 ))
do
    stuff
    sleep 42
done

然后，当您想在下一次迭代中退出循环时:

Then, when you want to exit the loop at the next iteration:

kill -SIGUSR1 pid

其中pid是脚本的进程ID.如果在睡眠期间发出信号，它将唤醒(睡眠进入睡眠状态，直到出现任何信号为止).

Where pid is the process-id of the script. If the signal is raised during the sleep, it will wake (sleep sleeps until any signal occurs).

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