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
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
}
其他回答
下面是一个使用锁文件并将PID回显到锁文件中的实现。如果进程在删除pidfile之前被杀死,这可以起到保护作用:
LOCKFILE=/tmp/lock.txt
if [ -e ${LOCKFILE} ] && kill -0 `cat ${LOCKFILE}`; then
echo "already running"
exit
fi
# make sure the lockfile is removed when we exit and then claim it
trap "rm -f ${LOCKFILE}; exit" INT TERM EXIT
echo $$ > ${LOCKFILE}
# do stuff
sleep 1000
rm -f ${LOCKFILE}
这里的技巧是kill -0,它不传递任何信号,只是检查具有给定PID的进程是否存在。此外,调用trap将确保即使进程被杀死(kill -9除外),锁文件也会被移除。
这个例子是在man flock中解释的,但它需要一些改进,因为我们应该管理bug和退出代码:
#!/bin/bash
#set -e this is useful only for very stupid scripts because script fails when anything command exits with status more than 0 !! without possibility for capture exit codes. not all commands exits >0 are failed.
( #start subprocess
# Wait for lock on /var/lock/.myscript.exclusivelock (fd 200) for 10 seconds
flock -x -w 10 200
if [ "$?" != "0" ]; then echo Cannot lock!; exit 1; fi
echo $$>>/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
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
使用进程的锁更强大,还可以处理不合理的退出。 只要进程在运行,Lock_file就保持打开状态。一旦进程存在,它将被关闭(通过shell)(即使它被杀死)。 我发现这个方法非常有效:
lock_file=/tmp/`basename $0`.lock
if fuser $lock_file > /dev/null 2>&1; then
echo "WARNING: Other instance of $(basename $0) running."
exit 1
fi
exec 3> $lock_file