Question

假设日志中有4种不同类型的模式（错误），每种模式都可能不时发生。例如：＆＃34;超时异常＆＃34;，＆＃34; ldap错误＆＃34;，＆＃34; db error＆＃34;，＆＃34;错误四＆＃34;。任何一个地方都可以为我提供一个脚本： - 如何每小时在日志中grep多个模式，如果脚本找到任何模式，那么它应该只向我发送一次警报，没有重复警报。请帮我。谢谢

Answer 1

#!/bin/bash

while true; do
    export ERRORS=`cat YOUR_LOG_FILE | grep -e "(timeout exception)|(ldap error)|(db error)|(error four)"
    if [ $ERRORS ]; then
        # sendmail or any other kind of "alert" you prefer.
        echo $ERRORS | sendmail "your@email.com"
    fi
    sleep 1h
done

Answer 2

创建一个每小时运行一次的crontab条目。该条目可以调用您的脚本：

logfile=/path/to/logfile/application.out

function send_alert {
   # Some sendmail or other tool to send your alert using the args
   printf "I want to alert about %s" "$*"
}

# Solution only announcing errors without sending them
grep -qE "timeout exception|ldap error|db error|error four" ${logfile} && 
        send_alert "grep found something"

# Solution sending number of errorlines
errorlinecount=$(grep -c "timeout exception|ldap error|db error|error four" )
if [ ${errorcount} -gt 0 ]; then
    send_alert "grep found ${errorcount} disturbing lines"
fi

Unix每小时grep多个模式

2 个答案: