确保Bash脚本只有一个实例正在运行的最佳方法是什么?

问题描述

确保给定脚本的仅一个实例正在运行的最简单/最佳方法是什么(假设它是Linux上的Bash)?

What is the simplest/best way to ensure only one instance of a given script is running - assuming it's Bash on Linux?

此刻我正在做

ps -C script.name.sh > /dev/null 2>&1 || ./script.name.sh

但是它有几个问题:

1. 将支票置于脚本之外
2. 不允许我从单独的帐户运行相同的脚本-有时我想这样做.
3. -C仅检查进程名称的前14个字符

1. it puts the check outside of script
2. it doesn't let me run the same script from separate accounts - which I would like sometimes.
3. -C checks only first 14 characters of process name

当然，我可以编写自己的pidfile处理程序，但我认为应该有一种简单的方法来实现它.

Of course, I can write my own pidfile handling, but I sense that there should be a simple way to do it.

推荐答案

如果所有用户的脚本都相同，则可以使用lockfile方法.如果您获得了锁，请继续执行其他操作，并显示一条消息并退出.

If the script is the same across all users, you can use a lockfile approach. If you acquire the lock, proceed else show a message and exit.

例如:

[Terminal #1] $ lockfile -r 0 /tmp/the.lock
[Terminal #1] $

[Terminal #2] $ lockfile -r 0 /tmp/the.lock
[Terminal #2] lockfile: Sorry, giving up on "/tmp/the.lock"

[Terminal #1] $ rm -f /tmp/the.lock
[Terminal #1] $

[Terminal #2] $ lockfile -r 0 /tmp/the.lock
[Terminal #2] $

获取/tmp/the.lock后，您的脚本将是唯一有权执行的脚本.完成后，只需取下锁.在脚本形式下，它可能类似于:

After /tmp/the.lock has been acquired your script will be the only one with access to execution. When you are done, just remove the lock. In script form this might look like:

#!/bin/bash

lockfile -r 0 /tmp/the.lock || exit 1

# Do stuff here

rm -f /tmp/the.lock

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