在bash脚本按顺序运行命令 [英] Sequentially run commands in bash script

问题描述

我需要运行顺序在守护进程模式两个命令（命令将输出stderr上错误）。问题是，即使我在倾倒的/ dev / null的所有的输出，第二个命令（run_cmd2）不能被调用。这里是我的脚本

I need to run two commands sequentially in daemon mode (the commands will output the errors on stderr). The problem is that even I dump all the output in /dev/null, the second command (run_cmd2) cannot be invoked. Here is my script

#! /bin/bash
nohup ./run_cmd1 &> /dev/null &
nohup ./run_cmd2 &> /dev/null &

任何想法？先谢谢你。

Any ideas? Thank you in advance.

推荐答案

如何使用文件进行通信的状态？

How about using a file to communicate state?

run_cmd2会等到一个文件存在运行前

run_cmd2 will wait until a file exists before running

当run_cmd1完成后，它会创建表示文件。

when run_cmd1 is done, it will create the said file.

在其运行run_cmd2将删除该文件的结尾，所以run_cmd1可以再次运行

at the end of its run the run_cmd2 will delete the file, so run_cmd1 can run again

或者他们使用端口相互交谈？

Or maybe they talk to each other using a port?

也许一个信号？

有关详细信息： http://www.tldp.org/LDP/ TLK / IPC / ipc.html

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