我想要做的应该是非常简单,我自己已经达到了下面的解决方案,我需要的是一些指示,告诉我这是否是这样做的方法或者我应该重构代码中的任何内容。
下面的代码应该创建一些并行进程并等待它们完成执行,然后一次又一次地重新运行代码......
脚本由cron作业在10分钟时触发一次,如果脚本正在运行,则不执行任何操作,否则启动工作进程。
由于我不熟悉bash编程,因此非常感谢任何见解。
#!/bin/bash
# paths
THISPATH="$( cd "$( dirname "$0" )" && pwd )"
# make sure we move in the working directory
cd $THISPATH
# console init path
CONSOLEPATH="$( cd ../../ && pwd )/console.php"
# command line arguments
daemon=0
PHPPATH="/usr/bin/php"
help=0
# flag for binary search
LOOKEDFORPHP=0
# arguments init
while getopts d:p:h: opt; do
case $opt in
d)
daemon=$OPTARG
;;
p)
PHPPATH=$OPTARG
LOOKEDFORPHP=1
;;
h)
help=$OPTARG
;;
esac
done
shift $((OPTIND - 1))
# allow only one process
processesLength=$(ps aux | grep -v "grep" | grep -c $THISPATH/send-campaigns-daemon.sh)
if [ ${processesLength:-0} -gt 2 ]; then
# The process is already running
exit 0
fi
if [ $help -eq 1 ]; then
echo "---------------------------------------------------------------"
echo "| Usage: send-campaigns-daemon.sh |"
echo "| To force PHP CLI binary : |"
echo "| send-campaigns-daemon.sh -p /path/to/php-cli/binary |"
echo "---------------------------------------------------------------"
exit 0
fi
# php executable path, find it if not provided
if [ $PHPPATH ] && [ ! -f $PHPPATH ] && [ $LOOKEDFORPHP -eq 0 ]; then
phpVariants=( "php-cli" "php5-cli" "php5" "php" )
LOOKEDFORPHP=1
for i in "${phpVariants[@]}"
do
which $i >/dev/null 2>&1
if [ $? -eq 0 ]; then
PHPPATH=$(which $i)
fi
done
fi
if [ ! $PHPPATH ] || [ ! -f $PHPPATH ]; then
# Did not find PHP
exit 1
fi
# load options from app
parallelProcessesPerCampaign=3
campaignsAtOnce=10
subscribersAtOnce=300
sleepTime=30
function loadOptions {
local COMMAND="$PHPPATH $CONSOLEPATH option get_option --name=%s --default=%d"
parallelProcessesPerCampaign=$(printf "$COMMAND" "system.cron.send_campaigns.parallel_processes_per_campaign" 3)
campaignsAtOnce=$(printf "$COMMAND" "system.cron.send_campaigns.campaigns_at_once" 10)
subscribersAtOnce=$(printf "$COMMAND" "system.cron.send_campaigns.subscribers_at_once" 300)
sleepTime=$(printf "$COMMAND" "system.cron.send_campaigns.pause" 30)
parallelProcessesPerCampaign=$($parallelProcessesPerCampaign)
campaignsAtOnce=$($campaignsAtOnce)
subscribersAtOnce=$($subscribersAtOnce)
sleepTime=$($sleepTime)
}
# define the daemon function that will stay in loop
function daemon {
loadOptions
local pids=()
local k=0
local i=0
local COMMAND="$PHPPATH -q $CONSOLEPATH send-campaigns --campaigns_offset=%d --campaigns_limit=%d --subscribers_offset=%d --subscribers_limit=%d --parallel_process_number=%d --parallel_processes_count=%d --usleep=%d --from_daemon=1"
while [ $i -lt $campaignsAtOnce ]
do
while [ $k -lt $parallelProcessesPerCampaign ]
do
parallelProcessNumber=$(( $k + 1 ))
usleep=$(( $k * 10 + $i * 10 ))
CMD=$(printf "$COMMAND" $i 1 $(( $subscribersAtOnce * $k )) $subscribersAtOnce $parallelProcessNumber $parallelProcessesPerCampaign $usleep)
$CMD > /dev/null 2>&1 &
pids+=($!)
k=$(( k + 1 ))
done
i=$(( i + 1 ))
done
waitForPids pids
sleep $sleepTime
daemon
}
function daemonize {
$THISPATH/send-campaigns-daemon.sh -d 1 -p $PHPPATH > /dev/null 2>&1 &
}
function waitForPids {
stillRunning=0
for i in "${pids[@]}"
do
if ps -p $i > /dev/null
then
stillRunning=1
break
fi
done
if [ $stillRunning -eq 1 ]; then
sleep 0.5
waitForPids pids
fi
return 0
}
if [ $daemon -eq 1 ]; then
daemon
else
daemonize
fi
exit 0
答案 0 :(得分:1)
启动脚本时,创建一个锁定文件以了解此脚本是否正在运行。脚本完成后,删除锁定文件。如果有人在运行过程中终止该进程,则锁定文件将永久保留,但测试它的大小,如果超过定义的值则删除。例如,
#!/bin/bash
# 10 min
LOCK_MAX=600
typedef LOCKFILE=/var/lock/${0##*/}.lock
if [[ -f $LOCKFILE ]] ; then
TIMEINI=$( stat -c %X $LOCKFILE )
SEGS=$(( $(date +%s) - $TIEMPOINI ))
if [[ $SEGS -gt $LOCK_MAX ]] ; then
reportLocking or somethig to inform you
# Kill old intance ???
OLDPID=$(<$LOCKFILE)
[[ -e /proc/$OLDPID ]] && kill -9 $OLDPID
# Next time that the program is run, there is no lock file and it will run.
rm $LOCKFILE
fi
exit 65
fi
# Save PID of this instance to the lock file
echo "$$" > $LOCKFILE
### Your code go here
# Remove the lock file before script finish
[[ -e $LOCKFILE ]] && rm $LOCKFILE
exit 0
答案 1 :(得分:1)
来自here:
#!/bin/bash
...
echo PARALLEL_JOBS:${PARALLEL_JOBS:=1}
declare -a tests=($(.../find_what_to_run))
echo "${tests[@]}" | \
xargs -d' ' -n1 -P${PARALLEL_JOBS} -I {} bash -c ".../run_that {}" || { echo "FAILURE"; exit 1; }
echo "SUCCESS"
和here您可以使用fuser
答案 2 :(得分:1)
好的,所以我想我可以回答我自己的问题,并在多次测试后使用正确的答案 所以这里是最终版本,简化,没有comments / echo:
#!/bin/bash
sleep 2
DIR="$( cd "$( dirname "$0" )" && pwd )"
FILE_NAME="$( basename "$0" )"
COMMAND_FILE_PATH="$DIR/$FILE_NAME"
if [ ! -f "$COMMAND_FILE_PATH" ]; then
exit 1
fi
cd $DIR
CONSOLE_PATH="$( cd ../../ && pwd )/console.php"
PHP_PATH="/usr/bin/php"
help=0
LOOKED_FOR_PHP=0
while getopts p:h: opt; do
case $opt in
p)
PHP_PATH=$OPTARG
LOOKED_FOR_PHP=1
;;
h)
help=$OPTARG
;;
esac
done
shift $((OPTIND - 1))
if [ $help -eq 1 ]; then
printf "%s\n" "HELP INFO"
exit 0
fi
if [ "$PHP_PATH" ] && [ ! -f "$PHP_PATH" ] && [ "$LOOKED_FOR_PHP" -eq 0 ]; then
php_variants=( "php-cli" "php5-cli" "php5" "php" )
LOOKED_FOR_PHP=1
for i in "${php_variants[@]}"
do
which $i >/dev/null 2>&1
if [ $? -eq 0 ]; then
PHP_PATH="$(which $i)"
break
fi
done
fi
if [ ! "$PHP_PATH" ] || [ ! -f "$PHP_PATH" ]; then
exit 1
fi
LOCK_BASE_PATH="$( cd ../../../common/runtime && pwd )/shell-pids"
LOCK_PATH="$LOCK_BASE_PATH/send-campaigns-daemon.pid"
function remove_lock {
if [ -d "$LOCK_PATH" ]; then
rmdir "$LOCK_PATH" > /dev/null 2>&1
fi
exit 0
}
if [ ! -d "$LOCK_BASE_PATH" ]; then
if ! mkdir -p "$LOCK_BASE_PATH" > /dev/null 2>&1; then
exit 1
fi
fi
process_running=0
if mkdir "$LOCK_PATH" > /dev/null 2>&1; then
process_running=0
else
process_running=1
fi
if [ $process_running -eq 1 ]; then
exit 0
fi
trap "remove_lock" 1 2 3 15
COMMAND="$PHP_PATH $CONSOLE_PATH option get_option --name=%s --default=%d"
parallel_processes_per_campaign=$(printf "$COMMAND" "system.cron.send_campaigns.parallel_processes_per_campaign" 3)
campaigns_at_once=$(printf "$COMMAND" "system.cron.send_campaigns.campaigns_at_once" 10)
subscribers_at_once=$(printf "$COMMAND" "system.cron.send_campaigns.subscribers_at_once" 300)
sleep_time=$(printf "$COMMAND" "system.cron.send_campaigns.pause" 30)
parallel_processes_per_campaign=$($parallel_processes_per_campaign)
campaigns_at_once=$($campaigns_at_once)
subscribers_at_once=$($subscribers_at_once)
sleep_time=$($sleep_time)
k=0
i=0
pp=0
COMMAND="$PHP_PATH -q $CONSOLE_PATH send-campaigns --campaigns_offset=%d --campaigns_limit=%d --subscribers_offset=%d --subscribers_limit=%d --parallel_process_number=%d --parallel_processes_count=%d --usleep=%d --from_daemon=1"
while [ $i -lt $campaigns_at_once ]
do
while [ $k -lt $parallel_processes_per_campaign ]
do
parallel_process_number=$(( $k + 1 ))
usleep=$(( $k * 10 + $i * 10 ))
CMD=$(printf "$COMMAND" $i 1 $(( $subscribers_at_once * $k )) $subscribers_at_once $parallel_process_number $parallel_processes_per_campaign $usleep)
$CMD > /dev/null 2>&1 &
k=$(( k + 1 ))
pp=$(( pp + 1 ))
done
i=$(( i + 1 ))
done
wait
sleep ${sleep_time:-30}
$COMMAND_FILE_PATH -p "$PHP_PATH" > /dev/null 2>&1 &
remove_lock
exit 0
答案 3 :(得分:1)
通常,它是一个锁定文件,而不是锁定路径。将PID保存在锁定文件中以监视进程。在这种情况下,您的锁定目录不包含任何PID信息。如果在不清除锁定的情况下不正确关闭进程,则脚本在启动时也不会执行任何PID文件/目录维护。
考虑到这一点,我更喜欢你的第一个脚本。直接监控PID运行更清晰。唯一的问题是如果你用cron启动第二个实例,它不知道PID连接到第一个实例。
你也有processLength -gt 2,它是2,而不是1个进程,所以你将复制你的进程线程。
似乎daemonize只是用守护进程调用脚本,这不是很有用。另外,使用与函数同名的变量无效。
答案 4 :(得分:1)
制作锁文件的正确方法如下:
# Create a temporary file
echo $$ > ${LOCKFILE}.tmp$$
# Try the lock; ln without -f is atomic
if ln ${LOCKFILE}.tmp$$ ${LOCKFILE}; then
# we got the lock
else
# we didn't get the lock
fi
# Tidy up the temporary file
rm ${LOCKFILE}.tmp$$
释放锁定:
# Unlock
rm ${LOCKFILE}
关键是使用唯一名称将锁文件创建到一侧,然后尝试将其链接到真实名称。这是一个原子操作,所以它应该是安全的。
任何“测试和设置”的解决方案都会为您提供竞争条件。是的,可以整理出来,但最终会编写额外的代码。