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

一个有flock(1)但没有subshell的例子。Flock ()ed文件/tmp/foo永远不会被删除,但这没关系,因为它会被Flock()和un-flock()ed。

#!/bin/bash

exec 9<> /tmp/foo
flock -n 9
RET=$?
if [[ $RET -ne 0 ]] ; then
    echo "lock failed, exiting"
    exit
fi

#Now we are inside the "critical section"
echo "inside lock"
sleep 5
exec 9>&- #close fd 9, and release lock

#The part below is outside the critical section (the lock)
echo "lock released"
sleep 5

其他回答

对于shell脚本,我倾向于使用mkdir而不是flock,因为它使锁更可移植。

不管怎样,使用set -e是不够的。它只在任何命令失败时退出脚本。你的锁还是会留下的。

为了正确的锁清理,你真的应该把你的陷阱设置成这样的伪代码(提取,简化和未经测试,但来自积极使用的脚本):

#=======================================================================
# Predefined Global Variables
#=======================================================================

TMPDIR=/tmp/myapp
[[ ! -d $TMP_DIR ]] \
    && mkdir -p $TMP_DIR \
    && chmod 700 $TMPDIR

LOCK_DIR=$TMP_DIR/lock

#=======================================================================
# Functions
#=======================================================================

function mklock {
    __lockdir="$LOCK_DIR/$(date +%s.%N).$$" # Private Global. Use Epoch.Nano.PID

    # If it can create $LOCK_DIR then no other instance is running
    if $(mkdir $LOCK_DIR)
    then
        mkdir $__lockdir  # create this instance's specific lock in queue
        LOCK_EXISTS=true  # Global
    else
        echo "FATAL: Lock already exists. Another copy is running or manually lock clean up required."
        exit 1001  # Or work out some sleep_while_execution_lock elsewhere
    fi
}

function rmlock {
    [[ ! -d $__lockdir ]] \
        && echo "WARNING: Lock is missing. $__lockdir does not exist" \
        || rmdir $__lockdir
}

#-----------------------------------------------------------------------
# Private Signal Traps Functions {{{2
#
# DANGER: SIGKILL cannot be trapped. So, try not to `kill -9 PID` or 
#         there will be *NO CLEAN UP*. You'll have to manually remove 
#         any locks in place.
#-----------------------------------------------------------------------
function __sig_exit {

    # Place your clean up logic here 

    # Remove the LOCK
    [[ -n $LOCK_EXISTS ]] && rmlock
}

function __sig_int {
    echo "WARNING: SIGINT caught"    
    exit 1002
}

function __sig_quit {
    echo "SIGQUIT caught"
    exit 1003
}

function __sig_term {
    echo "WARNING: SIGTERM caught"    
    exit 1015
}

#=======================================================================
# Main
#=======================================================================

# Set TRAPs
trap __sig_exit EXIT    # SIGEXIT
trap __sig_int INT      # SIGINT
trap __sig_quit QUIT    # SIGQUIT
trap __sig_term TERM    # SIGTERM

mklock

# CODE

exit # No need for cleanup code here being in the __sig_exit trap function

接下来会发生什么。所有陷阱都会产生一个出口,所以__sig_exit函数总是会发生(除非SIGKILL),它会清理你的锁。

注意:我的退出值不是低值。为什么?各种批处理系统生成或期望数字0到31。将它们设置为其他内容,我可以让我的脚本和批处理流对前一个批处理作业或脚本做出相应的反应。

如果您不想或不能使用flock(例如,您没有使用共享文件系统),请考虑使用外部服务,如lockable。

它暴露了咨询锁原语,就像flock一样。特别地,你可以通过以下方式获取锁:

https://lockable.dev/api/acquire/my-lock-name

然后通过

https://lockable.dev/api/release/my-lock-name

通过将脚本执行与锁获取和释放结合在一起,您可以确保在任何给定时间只有一个流程实例在运行。

如果flock的限制,这已经在这篇文章的其他地方描述过了,对你来说不是问题,那么这应该是有效的:

#!/bin/bash

{
    # exit if we are unable to obtain a lock; this would happen if 
    # the script is already running elsewhere
    # note: -x (exclusive) is the default
    flock -n 100 || exit

    # put commands to run here
    sleep 100
} 100>/tmp/myjob.lock 

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

我使用onlineer @脚本的开头:

#!/bin/bash

if [[ $(pgrep -afc "$(basename "$0")") -gt "1" ]]; then echo "Another instance of "$0" has already been started!" && exit; fi
.
the_beginning_of_actual_script

在内存中看到进程的存在是很好的(不管进程的状态是什么);但它对我很有用。