将bash脚本作为守护进程运行 [英] Run bash script as daemon

问题描述

我有一个脚本，每次运行我的PHP脚本X次：

I have a script, which runs my PHP script each X times:

#!/bin/bash
while true; do
/usr/bin/php -f ./my-script.php
echo "Waiting..."
sleep 3
done

如何以守护进程启动它？

How can I start it as daemon?

推荐答案

要从shell运行它作为完整的守护进程，你需要使用 setsid 并重定向它的输出。您可以将输出重定向到日志文件，或 / dev / null 以将其丢弃。假设您的脚本被称为myscript.sh，使用以下命令：

To run it as a full daemon from a shell, you'll need to use setsid and redirect its output. You can redirect the output to a logfile, or to /dev/null to discard it. Assuming your script is called myscript.sh, use the following command:

setsid myscript.sh >/dev/null 2>&1 < /dev/null &

这将完全从当前shell（stdin，stdout和stderr）中分离进程。如果要将输出保存在日志文件中，请将第一个 / dev / null 替换为/ path / to / logfile。

This will completely detach the process from your current shell (stdin, stdout and stderr). If you want to keep the output in a logfile, replace the first /dev/null with your /path/to/logfile.

您必须重定向输出，否则它不会作为真正的守护进程运行（它将取决于您的shell读写输出）。

You have to redirect the output, otherwise it will not run as a true daemon (it will depend on your shell to read and write output).

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