![如何在 Redhat 下将 shell 脚本作为守护进程运行?](https://linux22.com/image/31266/%E5%A6%82%E4%BD%95%E5%9C%A8%20Redhat%20%E4%B8%8B%E5%B0%86%20shell%20%E8%84%9A%E6%9C%AC%E4%BD%9C%E4%B8%BA%E5%AE%88%E6%8A%A4%E8%BF%9B%E7%A8%8B%E8%BF%90%E8%A1%8C%EF%BC%9F.png)
我有一个 shell 脚本,它本质上是一个带有一些日志记录的单行代码,我试图从 init 脚本运行它。我正在使用daemon
其中的函数/etc/init.d/functions
来运行它,因为 Redhat 似乎不可start-stop-daemon
用。当我调用 init 脚本 ( /etc/init.d/script start
) 时,它停留在前台,而不是完成并让进程保持运行。我让这个脚本守护进程的正确方法是什么?
要运行的脚本:
# conf file where variables are defined
. /etc/script.conf
echo "Starting..." | logger -i
echo "Monitoring $LOG_LOCATION." | logger -i
echo "Sending to $MONITOR_HOST:$MONITOR_PORT." | logger -i
tail -n 1 -F $LOG_LOCATION |
grep WARN --line-buffered |
/usr/bin/nc -vv $MONITOR_HOST $MONITOR_PORT 2>&1 |
logger -i
初始化脚本:
#!/bin/bash
# Source Defaults
. /etc/default/script
# Source init functions
. /etc/init.d/functions
prog=/usr/local/bin/script.sh
[ -f /etc/script.conf ] || exit 1
RETVAL=0
start()
{
# Quit if disabled
if ! $ENABLED; then
echo "Service Disabled in /etc/default/script"
exit 1
fi
echo "Starting $prog"
daemon $prog
RETVAL=$?
return $RETVAL
}
stop ()
{
echo -n $"Stopping $prog: "
killproc $prog
RETVAL=$?
return $RETVAL
}
reload()
{
echo "Reload command is not implemented for this service."
return $RETVAL
}
restart()
{
stop
start
}
condrestart()
{
echo "Not Implemented."
}
# See how we were called.
case "$1" in
start)
start
;;
stop)
stop
;;
status)
status $prog
;;
restart)
restart
;;
reload)
reload
;;
condrestart)
condrestart
;;
*)
echo $"Usage: $0 {start|stop|status|restart|condrestart|reload}"
RETVAL=1
esac
使用 bash -vx 执行最后约 20 行:
+ case "$1" in
+ start
+ true
+ echo 'Starting /usr/local/bin/script.sh'
Starting /usr/local/bin/script.sh
+ daemon /usr/local/bin/script.sh
+ local gotbase= force=
+ local base= user= nice= bg= pid=
+ nicelevel=0
+ '[' /usr/local/bin/script.sh '!=' /usr/local/bin/script.sh ']'
+ '[' -z '' ']'
+ base=script.sh
+ '[' -f /var/run/script.sh.pid ']'
+ '[' -n '' -a -z '' ']'
+ ulimit -S -c 0
+ '[' -n '' ']'
+ '[' color = verbose -a -z '' ']'
+ '[' -z '' ']'
+ initlog -q -c /usr/local/bin/script.sh
答案1
我在以下位置找到了一个脚本http://www.linuxforums.org/forum/programming-scripting/190279-daemon-etc-init-d-functions-does-not-return-launching-process.html#post897522我可以对其进行修改以满足我的需要。它手动跟踪 PID 并使用pidof
.我最终不得不修改它来使用,pgrep
因为pidof
无法看到我的脚本的 PID。修改后,效果很好。 *注意,pgrep 似乎仅在完整脚本名称长度低于 15 个字符时才有效
这就是我最终得到的结果:
#!/bin/bash
#
#
#
# Start on runlevels 3, 4 and 5. Start late, kill early.
# chkconfig: 345 95 05
#
#
#!/bin/bash
# absolute path to executable binary
progpath='/usr/local/bin/script.sh'
# arguments to script
opts=''
# binary program name
prog=$(basename $progpath)
# pid file
pidfile="/var/run/${prog}.pid"
# make sure full path to executable binary is found
! [ -x $progpath ] && echo "$progpath: executable not found" && exit 1
eval_cmd() {
local rc=$1
if [ $rc -eq 0 ]; then
echo '[ OK ]'
else
echo '[FAILED]'
fi
return $rc
}
start() {
# see if running
local pids=$(pgrep $prog)
if [ -n "$pids" ]; then
echo "$prog (pid $pids) is already running"
return 0
fi
printf "%-50s%s" "Starting $prog: " ''
$progpath $opts &
# save pid to file if you want
echo $! > $pidfile
# check again if running
pgrep $prog >/dev/null 2>&1
eval_cmd $?
}
stop() {
# see if running
local pids=$(pgrep $prog)
if [ -z "$pids" ]; then
echo "$prog not running"
return 0
fi
printf "%-50s%s" "Stopping $prog: " ''
rm -f $pidfile
kill -9 $pids
eval_cmd $?
}
status() {
# see if running
local pids=$(pgrep $prog)
if [ -n "$pids" ]; then
echo "$prog (pid $pids) is running"
else
echo "$prog is stopped"
fi
}
case $1 in
start)
start
;;
stop)
stop
;;
status)
status
;;
restart)
stop
sleep 1
start
;;
*)
echo "Usage: $0 {start|stop|status|restart}"
exit 1
esac
exit $?
答案2
我不了解 Redhat,但daemon $prog &
对我来说看起来很奇怪。如果已经有一个用于守护进程的函数,为什么有必要(并且有用)将此函数本身放在后台?因此尝试不使用&
.