如何让bash脚本执行的命令的进程id？ [英] How to get the process id of command executed in bash script?

问题描述

我有一个剧本我想运行两个程序在同一时间，一个是C程序，另一种是请将CPULimit，我想在后台首先启动C程序＆放大器;然后拿到C程序的PID，并把它交给请将CPULimit这也将在后台运行＆放大器;

我想下面这样做，它只是启动的第一个程序和永远不会启动请将CPULimit。

另外，我在Arch Linux的使用systemd运行此作为启动脚本根。

 ＃！/斌/庆典/ myprogram＆安培;PID = $！请将CPULimit -z -p $ PID -l 75安培;退出0
 

解决方案

我觉得我有这个现在解决了，根据这个位置：<一href=\"http://stackoverflow.com/questions/7686989/running-bash-commands-in-the-background-without-printing-job-and-process-ids\">link我需要包装的命令是这样的（命令）创建子壳。

 ＃！/斌/庆典（mygprgram＆安培;）
mypid = $！
（请将CPULimit -z -p $ mypid -l 75安培;）退出0
 

I have a script i want to run 2 programs at the same time, One is a c program and the other is cpulimit, I want to start the C program in the background first with "&" and then get the PID of the C program and hand it to cpulimit which will also run in the background with "&".

I tried doing this below and it just starts the first program and never starts cpulimit.

Also i am running this as a startup script as root using systemd in arch linux.

#!/bin/bash

/myprogram &

PID=$!

cpulimit -z -p $PID -l 75 &

exit 0

解决方案

I think i have this solved now, According to this here: link I need to wrap the commands like this (command) to create a sub shell.

#!/bin/bash

(mygprgram &)
mypid=$!
(cpulimit -z -p $mypid -l 75 &)

exit 0

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