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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