Linux Shell 脚本监控 WebLogic 的运行状态、健康状态、打开的套接字数

经常看到有人通过检查 PID 或者端口号是否存在来判断 WebLogic 实例是否正常。

其实那种方法是不好的。因为在实际的线上环境里面，WebLogic 实例会因为很多种原因出现异常，比如内存溢出、过载、死锁等，这个时候它的 PID 或端口都没有问题，但实际上它已经无法正常工作了，需要通过重启 WebLogic 实例才能恢复正常。

一个更加靠谱的监控方法是，在脚本里面调用 WebLogic 的命令 weblogic.Admin 来判断实例的状态。希望本文的脚本对大家有所帮助。

操作系统版本：

[root@WLS1 /]# head -1 /etc/redhat-release
Red Hat Enterprise Linux Server release 5.3 (Tikanga)

WebLogic 版本：

[root@WLS1 /]# java weblogic.version | grep -m 1 "."
WebLogic Server 10.3.6.0  Tue Nov 15 08:52:36 PST 2011 1441050

代码：

#!/bin/bash
CLASSPATH="/opt/Oracle/Middleware/wlserver_10.3/server/lib/weblogic.jar:$CLASSPATH"
PATH="/usr/java/jdk1.6.0_45/bin:$PATH"

URL="192.168.222.11:7020"
USER_NAME="weblogic"
PASS_WORD="weblogic1"
DOMAIN_NAME="MedRecDomain"
SERVER_NAME="MedRecAdmSvr"

STATE_ALL=$(java weblogic.Admin -url $URL -username $USER_NAME -password $PASS_WORD get -pretty -mbean "$DOMAIN_NAME:Location=$SERVER_NAME,Name=$SERVER_NAME,Type=ServerRuntime")

# Check WebLogic instance running status
echo "$STATE_ALL" | grep -q "State: RUNNING"
if [ $? == 0 ]; then
    echo "$URL $DOMAIN_NAME $SERVER_NAME running status is OK"
else
    echo "$URL $DOMAIN_NAME $SERVER_NAME running status is not OK"
fi

# Check WebLogic instance health status
echo "$STATE_ALL" | grep -q "State:HEALTH_OK"
if [ $? == 0 ]; then
    echo "$URL $DOMAIN_NAME $SERVER_NAME health status is OK"
else
    echo "$URL $DOMAIN_NAME $SERVER_NAME health status is not OK"
fi

# Check WebLogic instance open sockets number
SOCKET_MAX=200
SOCKET_NOW=$(echo "$STATE_ALL" | awk '/OpenSocketsCurrentCount/{print $2}')
if [ x$SOCKET_NOW == x ]; then
    echo "$URL $DOMAIN_NAME $SERVER_NAME open sockets number is not OK: fail to get"
else
    if [ $SOCKET_NOW -gt $SOCKET_MAX ]; then
        echo "$URL $DOMAIN_NAME $SERVER_NAME health status is not OK: $SOCKET_NOW greater than $SOCKET_MAX"
    else
        echo "$URL $DOMAIN_NAME $SERVER_NAME health status is OK: $SOCKET_NOW not greater than $SOCKET_MAX"
    fi
fi