自动重启服务的shell脚本代码

2019-10-26 18:37:10

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来源：转载

供稿：网友

代码如下:
#!/bin/bash
if [ ! -f /tmp/down_count ];then
echo "0" > /tmp/down_count
fi
curl -I tomcat-host -o "/tmp/status" >/dev/null 2>&1
code=`awk 'NR==1 {print $2}' /tmp/status`
if [ "$[code]" -ge 500 ];then
down=`expr $(cat /tmp/down_count) + 1`
echo "$down" > /tmp/down_count
if [ "$down" -gt 3 ];then
if [ ! -f "/tmp/restart_count" ];then
echo "0" > /tmp/restart_count
fi
restart_count=`expr $(cat /tmp/restart_count) + 1`
echo "$restart_count" > /tmp/restart_count
if [ "$restart_count" -le 2 ];then
echo "tomcat down at `date`" >> /tmp/down_info
/etc/init.d/tomcat6 restart
fi
fi
else
echo "0" > /tmp/down_count
echo "0" > /tmp/restart_count
fi

脚本实现了，当检测网页状态码大于等于500连续出现3次数，自动重启tomcat6，且只连续重启两次。

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