1
我想要配置Tomcat的init.d啓動腳本的多個實例的工作(此時2個實例)啓動多個Tomcat實例
我按照下面的示例腳本創建的init.d腳本
#!/bin/bash
#
# tomcat This shell script takes care of starting and stopping Tomcat
#
# chkconfig: - 80 20
#
### BEGIN INIT INFO
# Provides: tomcat
# Required-Start: $network $syslog
# Required-Stop: $network $syslog
# Default-Start:
# Default-Stop:
# Short-Description: start and stop tomcat
### END INIT INFO
TOMCAT_USER=root
TOMCAT_HOME="/opt/tomcat7/node1"
SHUTDOWN_WAIT=45
tomcat_pid() {
echo `ps aux | grep org.apache.catalina.startup.Bootstrap | grep -v grep | awk '{ print $2 }'`
}
start() {
pid=$(tomcat_pid)
if [ -n "$pid" ]
then
echo "Tomcat is already running (pid: $pid)"
else
# Start tomcat
echo "Starting tomcat service"
/bin/su - -c "cd $TOMCAT_HOME/bin && $TOMCAT_HOME/bin/startup.sh" $TOMCAT_USER
fi
return 0
}
stop() {
pid=$(tomcat_pid)
if [ -n "$pid" ]
then
echo "Stoping Tomcat"
/bin/su - -c "cd $TOMCAT_HOME/bin && $TOMCAT_HOME/bin/shutdown.sh" $TOMCAT_USER
let kwait=$SHUTDOWN_WAIT
count=0
count_by=5
until [ `ps -p $pid | grep -c $pid` = '0' ] || [ $count -gt $kwait ]
do
echo "Waiting for processes to exit. Timeout before we kill the pid: ${count}/${kwait}"
sleep $count_by
let count=$count+$count_by;
done
if [ $count -gt $kwait ]; then
echo "Killing processes which didn't stop after $SHUTDOWN_WAIT seconds"
kill -9 $pid
fi
else
echo "Tomcat is not running"
fi
return 0
}
case $1 in
start)
start
;;
stop)
stop
;;
restart)
stop
start
;;
status)
pid=$(tomcat_pid)
if [ -n "$pid" ]
then
echo "Tomcat is running with pid: $pid"
else
echo "Tomcat is not running"
fi
;;
esac
exit 0
問題是tomcat_pid()
所有Tomcat實例的方法返回的進程ID,正因爲如此,二審不能啓動。有沒有更好的方法來處理這個問題?