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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