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
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
其他回答
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 $$ | tail -1`
if (echo "$UUID" > "$LOCKFILE") 2>/dev/null; then
ACQUIRED="TRUE"
return 0
else
if [ -e $LOCKFILE ]; then
# We may be dealing with a stale lock file.
# Bring out the magnifying glass.
CURRENT_UUID_FROM_LOCKFILE=`cat $LOCKFILE`
CURRENT_PID_FROM_LOCKFILE=`cat $LOCKFILE | cut -f 1 -d " "`
CURRENT_UUID_FROM_PS=`ps -eo pid,ppid,lstart $CURRENT_PID_FROM_LOCKFILE | tail -1`
if [ "$CURRENT_UUID_FROM_LOCKFILE" == "$CURRENT_UUID_FROM_PS" ]; then
echo "Script already running with following identification: $CURRENT_UUID_FROM_LOCKFILE" >&2
return 1
else
# The process that created this lock file died an ungraceful death.
# Take ownership of the lock file.
echo "The process $CURRENT_UUID_FROM_LOCKFILE is no longer around. Taking ownership of $LOCKFILE"
release "FORCE"
if (echo "$UUID" > "$LOCKFILE") 2>/dev/null; then
ACQUIRED="TRUE"
return 0
else
echo "Cannot write to $LOCKFILE. Error." >&2
return 1
fi
fi
else
echo "Do you have write permissons to $LOCKFILE ?" >&2
return 1
fi
fi
}
# Removes the lock file only if this script created it ($ACQUIRED is set),
# OR, if we are removing a stale lock file (first parameter is "FORCE")
release () {
#Destroy lock file. Take no prisoners.
if [ "$ACQUIRED" ] || [ "$1" == "FORCE" ]; then
rm -f $LOCKFILE
fi
}
# Test code
# int main( int argc, const char* argv[] )
echo "Acquring lock."
acquire
if [ $? -eq 0 ]; then
echo "Acquired lock."
read -p "Press [Enter] key to release lock..."
release
echo "Released lock."
else
echo "Unable to acquire lock."
fi
PID和锁文件绝对是最可靠的。当您尝试运行程序时,它可以检查锁文件,如果它存在,它可以使用ps查看进程是否仍在运行。如果不是,脚本可以启动,将锁文件中的PID更新为自己的PID。
已经回答了一百万次了,但是另一种方式,不需要外部依赖:
LOCK_FILE="/var/lock/$(basename "$0").pid"
trap "rm -f ${LOCK_FILE}; exit" INT TERM EXIT
if [[ -f $LOCK_FILE && -d /proc/`cat $LOCK_FILE` ]]; then
// Process already exists
exit 1
fi
echo $$ > $LOCK_FILE
每次它将当前PID($$)写入锁文件,并在脚本启动时检查进程是否正在使用最新的PID运行。
这将工作,如果你的脚本名称是唯一的:
#!/bin/bash
if [ $(pgrep -c $(basename $0)) -gt 1 ]; then
echo $(basename $0) is already running
exit 0
fi
如果scriptname不是唯一的,这在大多数linux发行版上都有效:
#!/bin/bash
exec 9>/tmp/my_lock_file
if ! flock -n 9 ; then
echo "another instance of this script is already running";
exit 1
fi
来源:http://mywiki.wooledge.org/BashFAQ/045
对于shell脚本,我倾向于使用mkdir而不是flock,因为它使锁更可移植。
不管怎样,使用set -e是不够的。它只在任何命令失败时退出脚本。你的锁还是会留下的。
为了正确的锁清理,你真的应该把你的陷阱设置成这样的伪代码(提取,简化和未经测试,但来自积极使用的脚本):
#=======================================================================
# Predefined Global Variables
#=======================================================================
TMPDIR=/tmp/myapp
[[ ! -d $TMP_DIR ]] \
&& mkdir -p $TMP_DIR \
&& chmod 700 $TMPDIR
LOCK_DIR=$TMP_DIR/lock
#=======================================================================
# Functions
#=======================================================================
function mklock {
__lockdir="$LOCK_DIR/$(date +%s.%N).$$" # 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。