bash Shell 脚本：如果脚本已经在运行，请确保不执行该脚本
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Shell script: Ensure that script isn't executed if already running
提问by Industrial
Possible Duplicate:
Quick-and-dirty way to ensure only one instance of a shell script is running at a time
I've set up a cronjob to backup my folders properly which I am quite proud of. However I've found out, by looking at the results from the backups, that my backup script has been called more than once by Crontab, resulting in multiple backups running at the same time.
我已经设置了一个 cronjob 来正确备份我的文件夹，我为此感到非常自豪。但是，通过查看备份结果，我发现我的备份脚本已被 Crontab 多次调用，导致同时运行多个备份。
Is there any way I can ensure that a certain shell script to not run if the very same script already is executing?
如果相同的脚本已经在执行，有什么方法可以确保某个 shell 脚本不会运行？
回答by Arnaud Le Blanc
A solution without race condition or early exit problems is to use a lock file. The flockutility handles this very well and can be used like this:
没有竞争条件或提前退出问题的解决方案是使用锁定文件。该flock实用程序可以很好地处理这个问题，可以像这样使用：
flock -n /var/run/your.lockfile -c /your/script
It will return immediately with a non 0 status if the script is already running.
如果脚本已经在运行，它将立即以非 0 状态返回。
回答by paxdiablo
The usual and simple way to do this is to put something like:
执行此操作的通常且简单的方法是输入以下内容：
if [[ -f /tmp/myscript.running ]] ; then
exit
fi
touch /tmp/myscript.running
at the top of you script and
在你的脚本顶部和
rm -f /tmp/myscript.running
at the end, and in trapfunctions in case it doesn't reach the end.
最后，在trap函数中，以防它没有到达终点。
This still has a few potential problems (such as a race condition at the top) but will do for the vast majority of cases.
这仍然有一些潜在的问题（例如顶部的竞争条件），但对于绝大多数情况都可以。
回答by David Hall
A good way without a lock file:
没有锁文件的好方法：
ps | grep ##代码## | grep -v grep > /var/tmp/##代码##.pid
pids=$(cat /var/tmp/##代码##.pid | cut -d ' ' -f 1)
for pid in $pids
do
if [ $pid -ne $$ ]; then
logprint " ##代码## is already running. Exiting"
exit 7
fi
done
rm -f /var/tmp/##代码##.pid
This does it without a lock file, which is cool. ps into a temp file, scrape the first field (pid #) and look for ourselves. If we find a different one, then somebody's already running. The grep $0 is to shorten the list to just those instances of this program, and the grep -v grep gets rid of the line that is the grep itself:)
这样做没有锁定文件，这很酷。ps 进入一个临时文件，抓取第一个字段（pid #）并查找我们自己。如果我们找到一个不同的，那么已经有人在跑了。grep $0 是将列表缩短为该程序的那些实例，而 grep -v grep 去掉了 grep 本身的行:)
回答by elp
You can use a tmp file.
Named it tmpCronBkp.a, tmpCronBkp.b, tmpCronBkp.c... etc. Releated of yout backup script.
您可以使用 tmp 文件。
将其命名为 tmpCronBkp.a、tmpCronBkp.b、tmpCronBkp.c...等。与您的备份脚本相关。
Create it on script start and delete it at the end...
在脚本开始时创建它并在最后删除它...
In a while, check if file exist or not and what file exist.
过一会儿，检查文件是否存在以及存在什么文件。
Have you tried this way?
你试过这种方法吗？
