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
看看FLOM (Free LOck Manager) http://sourceforge.net/projects/flom/:,您可以使用文件系统中不需要锁文件的抽象资源来同步命令和/或脚本。您可以在没有NFS(网络文件系统)服务器这样的NAS(网络附加存储)的情况下同步在不同系统中运行的命令。
使用最简单的用例,序列化“command1”和“command2”可能和执行一样简单:
flom -- command1
and
flom -- command2
来自两个不同的shell脚本。
其他回答
当目标是Debian机器时,我发现lockfile-progs包是一个很好的解决方案。Procmail还附带了一个锁文件工具。然而,有时这两种情况我都无法解决。
下面是我的解决方案,它使用mkdir来检测原子性,并使用PID文件来检测过期的锁。这段代码目前在Cygwin安装环境中运行,运行良好。
要使用它,当您需要独占访问某些东西时,只需调用exclusive_lock_require。一个可选的锁名参数允许您在不同的脚本之间共享锁。如果需要更复杂的功能,还有两个较低级别的函数(exclusive_lock_try和exclusive_lock_retry)。
function exclusive_lock_try() # [lockname]
{
local LOCK_NAME="${1:-`basename $0`}"
LOCK_DIR="/tmp/.${LOCK_NAME}.lock"
local LOCK_PID_FILE="${LOCK_DIR}/${LOCK_NAME}.pid"
if [ -e "$LOCK_DIR" ]
then
local LOCK_PID="`cat "$LOCK_PID_FILE" 2> /dev/null`"
if [ ! -z "$LOCK_PID" ] && kill -0 "$LOCK_PID" 2> /dev/null
then
# locked by non-dead process
echo "\"$LOCK_NAME\" lock currently held by PID $LOCK_PID"
return 1
else
# orphaned lock, take it over
( echo $$ > "$LOCK_PID_FILE" ) 2> /dev/null && local LOCK_PID="$$"
fi
fi
if [ "`trap -p EXIT`" != "" ]
then
# already have an EXIT trap
echo "Cannot get lock, already have an EXIT trap"
return 1
fi
if [ "$LOCK_PID" != "$$" ] &&
! ( umask 077 && mkdir "$LOCK_DIR" && umask 177 && echo $$ > "$LOCK_PID_FILE" ) 2> /dev/null
then
local LOCK_PID="`cat "$LOCK_PID_FILE" 2> /dev/null`"
# unable to acquire lock, new process got in first
echo "\"$LOCK_NAME\" lock currently held by PID $LOCK_PID"
return 1
fi
trap "/bin/rm -rf \"$LOCK_DIR\"; exit;" EXIT
return 0 # got lock
}
function exclusive_lock_retry() # [lockname] [retries] [delay]
{
local LOCK_NAME="$1"
local MAX_TRIES="${2:-5}"
local DELAY="${3:-2}"
local TRIES=0
local LOCK_RETVAL
while [ "$TRIES" -lt "$MAX_TRIES" ]
do
if [ "$TRIES" -gt 0 ]
then
sleep "$DELAY"
fi
local TRIES=$(( $TRIES + 1 ))
if [ "$TRIES" -lt "$MAX_TRIES" ]
then
exclusive_lock_try "$LOCK_NAME" > /dev/null
else
exclusive_lock_try "$LOCK_NAME"
fi
LOCK_RETVAL="${PIPESTATUS[0]}"
if [ "$LOCK_RETVAL" -eq 0 ]
then
return 0
fi
done
return "$LOCK_RETVAL"
}
function exclusive_lock_require() # [lockname] [retries] [delay]
{
if ! exclusive_lock_retry "$@"
then
exit 1
fi
}
我对现有的答案有以下问题:
Some answers try to clean up lock files and then having to deal with stale lock files caused by e.g. sudden crash/reboot. IMO that is unnecessarily complicated. Let lock files stay. Some answers use script file itself $0 or $BASH_SOURCE for locking often referring to examples from man flock. This fails when script is replaced due to update or edit causing next run to open and obtain lock on the new script file even though another instance holding a lock on the removed file is still running. Few answers use a fixed file descriptor. This is not ideal. I do not want to rely on how this will behave e.g. opening lock file fails but gets mishandled and attempts to lock on unrelated file descriptor inherited from parent process. Another fail case is injecting locking wrapper for a 3rd party binary that does not handle locking itself but fixed file descriptors can interfere with file descriptor passing to child processes. I reject answers using process lookup for already running script name. There are several reasons for it, such as but not limited to reliability/atomicity, parsing output, and having script that does several related functions some of which do not require locking.
这个答案是:
rely on flock because it gets kernel to provide locking ... provided lock file is created atomically and not replaced. assume and rely on lock file being stored on the local filesystem as opposed to NFS. change lock file presence to NOT mean anything about a running instance. Its role is purely to prevent two concurrent instances creating file with same name and replacing another's copy. Lock file does not get deleted, it gets left behind and can survive across reboots. The locking is indicated via flock not via lock file presence. assume bash shell, as tagged by the question.
它不是一个联机程序,但是没有注释和错误消息,它足够小:
#!/bin/bash
LOCKFILE=/var/lock/TODO
set -o noclobber
exec {lockfd}<> "${LOCKFILE}" || exit 1
set +o noclobber # depends on what you need
flock --exclusive --nonblock ${lockfd} || exit 1
但我更喜欢注释和错误消息:
#!/bin/bash
# TODO Set a lock file name
LOCKFILE=/var/lock/myprogram.lock
# Set noclobber option to ensure lock file is not REPLACED.
set -o noclobber
# Open lock file for R+W on a new file descriptor
# and assign the new file descriptor to "lockfd" variable.
# This does NOT obtain a lock but ensures the file exists and opens it.
exec {lockfd}<> "${LOCKFILE}" || {
echo "pid=$$ failed to open LOCKFILE='${LOCKFILE}'" 1>&2
exit 1
}
# TODO!!!! undo/set the desired noclobber value for the remainder of the script
set +o noclobber
# Lock on the allocated file descriptor or fail
# Adjust flock options e.g. --noblock as needed
flock --exclusive --nonblock ${lockfd} || {
echo "pid=$$ failed to obtain lock fd='${lockfd}' LOCKFILE='${LOCKFILE}'" 1>&2
exit 1
}
# DO work here
echo "pid=$$ obtained exclusive lock fd='${lockfd}' LOCKFILE='${LOCKFILE}'"
# Can unlock after critical section and do more work after unlocking
#flock -u ${lockfd};
# if unlocking then might as well close lockfd too
#exec {lockfd}<&-
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
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
试试下面的方法,
ab=`ps -ef | grep -v grep | grep -wc processname`
然后使用if循环将变量与1匹配。