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
我使用一种简单的方法来处理过期的锁文件。
注意,上面的一些解决方案存储pid,忽略了pid可以环绕的事实。因此,仅仅检查是否有一个有效的进程与存储的pid是不够的,特别是对于长时间运行的脚本。
我使用noclobber来确保一次只能打开一个脚本并写入锁文件。此外,我在锁文件中存储了足够的信息来惟一地标识一个进程。我定义了一组数据来唯一地标识一个进程为pid、ppid、lstart。
当一个新脚本启动时,如果它未能创建锁文件,那么它将验证创建锁文件的进程是否仍然存在。如果不是,我们假设原始进程不体面地死亡,并留下一个过时的锁文件。然后,新脚本获得锁文件的所有权,一切又恢复正常了。
应该与跨多个平台的多个shell一起工作。快速、便携、简单。
#!/usr/bin/env sh
# Author: rouble
LOCKFILE=/var/tmp/lockfile #customize this line
trap release INT TERM EXIT
# Creates a lockfile. Sets global variable $ACQUIRED to true on success.
#
# Returns 0 if it is successfully able to create lockfile.
acquire () {
set -C #Shell noclobber option. If file exists, > will fail.
UUID=`ps -eo pid,ppid,lstart $$ | tail -1`
if (echo "$UUID" > "$LOCKFILE") 2>/dev/null; then
ACQUIRED="TRUE"
return 0
else
if [ -e $LOCKFILE ]; then
# We may be dealing with a stale lock file.
# Bring out the magnifying glass.
CURRENT_UUID_FROM_LOCKFILE=`cat $LOCKFILE`
CURRENT_PID_FROM_LOCKFILE=`cat $LOCKFILE | cut -f 1 -d " "`
CURRENT_UUID_FROM_PS=`ps -eo pid,ppid,lstart $CURRENT_PID_FROM_LOCKFILE | tail -1`
if [ "$CURRENT_UUID_FROM_LOCKFILE" == "$CURRENT_UUID_FROM_PS" ]; then
echo "Script already running with following identification: $CURRENT_UUID_FROM_LOCKFILE" >&2
return 1
else
# The process that created this lock file died an ungraceful death.
# Take ownership of the lock file.
echo "The process $CURRENT_UUID_FROM_LOCKFILE is no longer around. Taking ownership of $LOCKFILE"
release "FORCE"
if (echo "$UUID" > "$LOCKFILE") 2>/dev/null; then
ACQUIRED="TRUE"
return 0
else
echo "Cannot write to $LOCKFILE. Error." >&2
return 1
fi
fi
else
echo "Do you have write permissons to $LOCKFILE ?" >&2
return 1
fi
fi
}
# Removes the lock file only if this script created it ($ACQUIRED is set),
# OR, if we are removing a stale lock file (first parameter is "FORCE")
release () {
#Destroy lock file. Take no prisoners.
if [ "$ACQUIRED" ] || [ "$1" == "FORCE" ]; then
rm -f $LOCKFILE
fi
}
# Test code
# int main( int argc, const char* argv[] )
echo "Acquring lock."
acquire
if [ $? -eq 0 ]; then
echo "Acquired lock."
read -p "Press [Enter] key to release lock..."
release
echo "Released lock."
else
echo "Unable to acquire lock."
fi
其他回答
PID和锁文件绝对是最可靠的。当您尝试运行程序时,它可以检查锁文件,如果它存在,它可以使用ps查看进程是否仍在运行。如果不是,脚本可以启动,将锁文件中的PID更新为自己的PID。
在脚本的开头添加这一行
[ "${FLOCKER}" != "$0" ] && exec env FLOCKER="$0" flock -en "$0" "$0" "$@" || :
这是人类群体的样板代码。
如果需要更多的日志记录,可以使用这个
[ "${FLOCKER}" != "$0" ] && { echo "Trying to start build from queue... "; exec bash -c "FLOCKER='$0' flock -E $E_LOCKED -en '$0' '$0' '$@' || if [ \"\$?\" -eq $E_LOCKED ]; then echo 'Locked.'; fi"; } || echo "Lock is free. Completing."
使用flock工具设置和检查锁。 这段代码通过检查FLOCKER变量来检测它是否第一次运行,如果它没有设置为脚本名称,那么它会尝试再次递归地使用flock启动脚本,并初始化FLOCKER变量,如果FLOCKER设置正确,那么在前一次迭代中flock成功,可以继续。如果锁繁忙,它将失败,并使用可配置的退出代码。
它似乎不能在Debian 7上工作,但似乎可以在实验util-linux 2.25包上再次工作。上面写着“羊群:……文本文件繁忙”。可以通过禁用脚本上的写权限来覆盖它。
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
你需要一个原子操作,比如flock,否则最终会失败。
但是如果没有羊群,该怎么办呢?这是mkdir。这也是一个原子操作。只有一个进程会成功执行mkdir,其他进程都会失败。
所以代码是:
if mkdir /var/lock/.myscript.exclusivelock
then
# do stuff
:
rmdir /var/lock/.myscript.exclusivelock
fi
你需要处理陈旧的锁,否则崩溃后你的脚本将永远不会再次运行。
这个例子是在man flock中解释的,但它需要一些改进,因为我们应该管理bug和退出代码:
#!/bin/bash
#set -e this is useful only for very stupid scripts because script fails when anything command exits with status more than 0 !! without possibility for capture exit codes. not all commands exits >0 are failed.
( #start subprocess
# Wait for lock on /var/lock/.myscript.exclusivelock (fd 200) for 10 seconds
flock -x -w 10 200
if [ "$?" != "0" ]; then echo Cannot lock!; exit 1; fi
echo $$>>/var/lock/.myscript.exclusivelock #for backward lockdir compatibility, notice this command is executed AFTER command bottom ) 200>/var/lock/.myscript.exclusivelock.
# Do stuff
# you can properly manage exit codes with multiple command and process algorithm.
# I suggest throw this all to external procedure than can properly handle exit X commands
) 200>/var/lock/.myscript.exclusivelock #exit subprocess
FLOCKEXIT=$? #save exitcode status
#do some finish commands
exit $FLOCKEXIT #return properly exitcode, may be usefull inside external scripts
你可以用另一种方法,列出我过去用过的过程。但这比上面的方法要复杂得多。你应该按ps列出进程,按其名称过滤,附加过滤器grep -v grep清除寄生虫,最后按grep -c计数。和数字比较。这是复杂而不确定的