有什么快速而简单的方法可以确保在给定时间内只有一个shell脚本实例在运行?
当前回答
下面是一个更优雅、更安全、更快速、更脏的方法,结合了上面提供的答案。
使用
包括sh_lock_functions.sh 使用sh_lock_init初始化 使用sh_acquire_lock进行锁定 使用sh_check_lock检查锁 使用sh_remove_lock解锁
脚本文件
sh_lock_functions.sh
#!/bin/bash
function sh_lock_init {
sh_lock_scriptName=$(basename $0)
sh_lock_dir="/tmp/${sh_lock_scriptName}.lock" #lock directory
sh_lock_file="${sh_lock_dir}/lockPid.txt" #lock file
}
function sh_acquire_lock {
if mkdir $sh_lock_dir 2>/dev/null; then #check for lock
echo "$sh_lock_scriptName lock acquired successfully.">&2
touch $sh_lock_file
echo $$ > $sh_lock_file # set current pid in lockFile
return 0
else
touch $sh_lock_file
read sh_lock_lastPID < $sh_lock_file
if [ ! -z "$sh_lock_lastPID" -a -d /proc/$sh_lock_lastPID ]; then # if lastPID is not null and a process with that pid exists
echo "$sh_lock_scriptName is already running.">&2
return 1
else
echo "$sh_lock_scriptName stopped during execution, reacquiring lock.">&2
echo $$ > $sh_lock_file # set current pid in lockFile
return 2
fi
fi
return 0
}
function sh_check_lock {
[[ ! -f $sh_lock_file ]] && echo "$sh_lock_scriptName lock file removed.">&2 && return 1
read sh_lock_lastPID < $sh_lock_file
[[ $sh_lock_lastPID -ne $$ ]] && echo "$sh_lock_scriptName lock file pid has changed.">&2 && return 2
echo "$sh_lock_scriptName lock still in place.">&2
return 0
}
function sh_remove_lock {
rm -r $sh_lock_dir
}
使用的例子
sh_lock_usage_example.sh
#!/bin/bash
. /path/to/sh_lock_functions.sh # load sh lock functions
sh_lock_init || exit $?
sh_acquire_lock
lockStatus=$?
[[ $lockStatus -eq 1 ]] && exit $lockStatus
[[ $lockStatus -eq 2 ]] && echo "lock is set, do some resume from crash procedures";
#monitoring example
cnt=0
while sh_check_lock # loop while lock is in place
do
echo "$sh_scriptName running (pid $$)"
sleep 1
let cnt++
[[ $cnt -gt 5 ]] && break
done
#remove lock when process finished
sh_remove_lock || exit $?
exit 0
特性
使用文件、目录和进程id的组合来锁定,以确保进程尚未运行 您可以检测脚本是否在删除锁之前停止(例如。进程终止,关闭,错误等) 您可以检查锁文件,并在丢失锁时使用它来触发进程关闭 Verbose,输出错误消息,以便于调试
其他回答
实际上,尽管bmdhacks的答案几乎很好,但在第一次检查锁文件之后和在写入它之前,第二个脚本有轻微的机会运行。它们都将写入锁文件,它们都将运行。以下是如何确保它有效的方法:
lockfile=/var/lock/myscript.lock
if ( set -o noclobber; echo "$$" > "$lockfile") 2> /dev/null ; then
trap 'rm -f "$lockfile"; exit $?' INT TERM EXIT
else
# or you can decide to skip the "else" part if you want
echo "Another instance is already running!"
fi
如果文件已经存在,noclobber选项将确保重定向命令将失败。因此,重定向命令实际上是原子的—您用一个命令写入和检查文件。你不需要在文件的末尾删除锁文件-它会被陷阱删除。我希望这对以后读到它的人有帮助。
另外,我没有看到Mikel已经正确地回答了这个问题,尽管他没有包括trap命令,以减少使用Ctrl-C停止脚本后留下锁文件的机会。这就是完整的解
我有一个基于文件名的简单解决方案
#!/bin/bash
MY_FILENAME=`basename "$BASH_SOURCE"`
MY_PROCESS_COUNT=$(ps a -o pid,cmd | grep $MY_FILENAME | grep -v grep | grep -v $$ | wc -
l)
if [ $MY_PROCESS_COUNT -ne 0 ]; then
echo found another process
exit 0
if
# Follows the code to get the job done.
上面有很多很好的答案。你也可以使用dotlockfile。
这是一些你可以在你的脚本中使用的示例代码:
$LOCKFILENAME=/var/run/test.lock
if ! dotlockfile -l -p -r 2 $LOCKFILENAME
then
echo "This test process already running!"
exit 1
fi
这将工作,如果你的脚本名称是唯一的:
#!/bin/bash
if [ $(pgrep -c $(basename $0)) -gt 1 ]; then
echo $(basename $0) is already running
exit 0
fi
如果scriptname不是唯一的,这在大多数linux发行版上都有效:
#!/bin/bash
exec 9>/tmp/my_lock_file
if ! flock -n 9 ; then
echo "another instance of this script is already running";
exit 1
fi
来源:http://mywiki.wooledge.org/BashFAQ/045
又快又脏?
#!/bin/sh
if [ -f sometempfile ]
echo "Already running... will now terminate."
exit
else
touch sometempfile
fi
..do what you want here..
rm sometempfile