如何向 bash 守护进程发送自定义信号？
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How to send custom signal to bash daemon process?
提问by greenV
I have simple bash daemon running (with root privileges ) in background which suppose to do action1or/and action2when notified.
我有一个简单的 bash 守护进程在后台运行（具有 root 权限），假设会action1或/和action2在通知时运行。
How do I notify it/send some kind of signal on which it will react?
我如何通知它/发送某种信号，它会做出反应？
I've tried scenarios with checking file change every 1 sec or more often, but that's kind of less-desirable solution.
我已经尝试过每 1 秒或更频繁地检查文件更改的场景，但这是一种不太理想的解决方案。
采纳答案by kasperd
You can send signals to a process using the killcommand. There is a range of standard signals as well as two user defined signals, which you can let your script handle whichever way you prefer. Here is how this could look in a script
您可以使用该kill命令向进程发送信号。有一系列标准信号以及两个用户定义的信号，您可以让脚本以您喜欢的任何方式处理它们。这是在脚本中的样子
#!/bin/bash
handler(){
echo "Handler was called"
}
trap handler USR1
while sleep 1
do
date
done
To send a signal to the script you first need to find the pid of the script and then use the killcommand. It could look like this kill -USR1 24962.
要向脚本发送信号，您首先需要找到脚本的 pid，然后使用该kill命令。它看起来像这样kill -USR1 24962。
回答by kasperd
You can use the killcommand to send a process a signal. In bash, you can use the trapcommand to create a signal handler.
您可以使用kill命令向进程发送信号。在 bash 中，您可以使用trap命令来创建信号处理程序。
#!/bin/bash
# traptest.sh
trap "echo Booh!" SIGINT SIGTERM
echo "pid is $$"
while : # This is the same as "while true".
do
sleep 60 # This script is not really doing anything.
done
