确保一次只运行一个shell脚本实例的快速方法

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

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。

2011-10-28 21:19:10

其他回答

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

2017-11-30 16:31:22

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。

2014-08-13 13:55:58

在flock(2)系统调用周围有一个包装器，毫无想象力地称为flock(1)。这使得可靠地获得独占锁相对容易，而不必担心清理等问题。手册页上有一些关于如何在shell脚本中使用它的示例。

2008-10-09 00:26:26

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

2012-05-03 18:16:49