Shell企業案例實戰和企業面試題

時間 2019-12-06

標籤 shell 企業案例實戰面試欄目 Unix 简体版

原文原文鏈接

shell企業面試題

一、批量建立帶有隨機小寫字符文件程序

使用for循環在/pizza目錄下建立10個html文件，其中每一個文件包含10個隨機小寫字母加固定字母_pizzahtml

一、思路分析：mysql

核心是：建立10個隨機小寫字母linux

第一種：$RANDOMweb

[root@web-01 /server/scripts]# echo $RANDOM
9839  範圍0-32767 ，第一個容易被破解，使用的時候最好再加個字符串

第二種：openssl rand -base64 10面試

[root@web-01 /server/scripts]# openssl rand -base64 10（最後面是長度）
yz+FH2zUNMlVnw==
[root@web-01 /server/scripts]# openssl rand -base64 100
wkfkVmliczOgoLl0z/m5S/7InZ8+4AzdHmR6t6hhE80oghRY46598L+no+HtDcHD
HyvQYnBWi6nQ0GbsjafyWZps7y6JpMEA6JOwQ+HlIOICXT7YLCcI9mQa6FUE+vHR
OcxHog==

第三種：datesql

[root@web-01 /server/scripts]# date +%N%s（）
3057895901553937109

第四種：head /dev/urandom |cksumshell

[root@web-01 /server/scripts]# head /dev/urandom |cksum
1831677657 1682

第五種：uuidgen數據庫

[root@web-01 /server/scripts]# uuidgen
218780c9-ee6f-41dc-9058-a9a3a717cde1

第六種：cat /proc/sys/kernel/random/uuid編程

[root@web-01 /server/scripts]# cat /proc/sys/kernel/random/uuid
542f71d9-b240-4891-b61d-632083ecf6be

第七種：expect的mkpasswdvim

[root@web-01 /server/scripts]# mkpasswd（yum install expect -y）
k3WlhJ|e7

[root@web-01 /server/scripts]# mkpasswd -l 20 -d 1 -c 2 （-l長度、-d數字、-c小寫字母、-C大寫字母、-s特殊字符）
nebiv;bnZi6vjluvczgP

本次使用$RANDOM，前面加字符串，md5加密，將數字替換爲字母，截取第2-第11個，共10位

[root@web-01 ~]# echo "PIZZA$RANDOM" |md5sum|tr "0-9" "a-z"|cut -c 2-11
befdbcdaee

二、for循環建立

path=/pizza
[ -d $path ] || mkdir $path
for n in {1..10}
do
    random=`echo "PIZZA$RANDOM" |md5sum|tr "0-9" "a-z"|cut -c 2-11`
    touch $path/${random}_pizza.html
done

二、批量更名

將題1中的pizza都改爲linux（最好用for循環實現，而且擴展名html所有改爲大寫）

思路分析：

一、先改一個

二、for循環

name=linux
path=/pizza/
cd $path
for file in `ls *.html`
do
    mv $file `echo ${file/pizza.html/linux.HTML}`                                   
done

三、方法二--用一條命令

[root@web-01 /pizza]# ls *.HTML |awk -F 'linux.HTML' '{print "mv",$0,$1"pizza.html"}' |bash

四、方法三--專業的rename

[root@web-01 /pizza]# rename "pizza.html" "linux.HTML" *.html

三、批量建立特殊需求要求用戶案例

批量建立10個系統帳號 pizza01-pizza10 並設置密碼（密碼是隨機數，要求字符和數字等混合）

思路分析：

一、建立10個帳號

第一種：echo pizza{01..10}

第二種：seq -w 10

二、隨機密碼，題1已經講了不少，此次用openssl rand -base64 100

三、建立用戶的命令

第一種：

useradd pizza01

echo 密碼 | passwd --stdin

第二種：chpasswd命令

格式要符合下面的形式

pizza01：passwd

pizza02：passwd

四、for循環

for n in {01..10}
do
    passwd=`openssl rand -base64 10`
    useradd pizza$n
    echo $passwd |passwd --stdin pizza$n
    echo -e "pizza$n\t$passwb" >> /pizza/user.list                                  
done

或者

for n in {01..10}
do
    passwd=`openssl rand -base64 10`
    useradd pizza$n
    echo "pizza$n:$passwd" >> /pizza/pass.log
done
chpasswd < /pizza/pass.log

五、優化，搞的專業一點

for n in {1..10}
do
    pass=`openssl rand -base64 10`                                                  
    if `grep "pizza$n" /etc/passwd &>/dev/null`    # 判斷是否是存在
    then
        useradd pizza$n &&\     # &&\設置強邏輯關係
            echo $pass|passwd --stdin pizza$n &&\
            echo -e "pizza$n\t$pass" >> /pizza/user.log
    else
        echo "pizza$n is exist."
    fi
done

優化：

for n in {1..10}
do
    pass=`openssl rand -base64 10`
    if [ `grep -w "pizza$n" /etc/passwd|wc -l` -eq 0 ]
    then
        useradd pizza$n &&\
            echo $pass|passwd --stdin pizza$n &&\
            echo -e "pizza$n\t$pass" >> /pizza/user.log
        echo "adduser successful"

    else
        echo "pizza$n is exist."
    fi
done

優化：

. /etc/init.d/functions
for n in {1..10}
do
    pass=`openssl rand -base64 10`
    if [ `grep -w "pizza$n" /etc/passwd|wc -l` -eq 0 ]
    then
        useradd pizza$n &&\
            echo $pass|passwd --stdin pizza$n &>/dev/null &&\
            echo -e "pizza$n\t$pass" >> /pizza/user.log
        action "adduser successful" /bin/true

    else
        action "pizza$n is exist." /bin/false
    fi
done

優化：

. /etc/init.d/functions
if [ $UID -ne 0 ]
then
    echo "必須用root執行本腳本"
    exit 1
fi

for n in {1..10}
do
    pass=`openssl rand -base64 10`
    if [ `grep -w "pizza$n" /etc/passwd|wc -l` -eq 0 ]
    then
        useradd pizza$n &&\
            echo $pass|passwd --stdin pizza$n &>/dev/null &&\
            echo -e "pizza$n\t$pass" >> /pizza/user.log
        action "adduser successful" /bin/true

    else
        action "pizza$n is exist." /bin/false
    fi
done

六、不用for循環的實現

http://user.qzone.qq.com/49000448/blog/1422183723

四、掃描網絡內存活主機

寫一個Shell腳本，判斷10.0.0.0/24網絡裏面，當前在線的IP有哪些

思路分析

一、判斷主機存活

ping  c 2 i 1 w 3 10.0.0.7
nmap  -sP 10.0.0.0/24

二、搞起來

第一種：ping

for n in {1..254}
do
    # 將整個執行用大括號括起來 加 &，進行批量ping，原理就是放到後臺執行
    {
    if `ping -c 1 -w 3 10.0.0.$n &>/dev/null`
    then
        echo "10.0.0.$n is up"
    else
        echo "10.0.0.$n is down"
    fi
} &
done

# 沒有進行過濾，因此輸出不少，能夠優化一下
第二種：使用nmap命令行就能夠

[root@web-01 /server/scripts]# nmap -sP 10.0.0.0/24 |awk '/Nmap scan report for/{print $NF}'

五、MySQL分庫備份

實現對MySQL數據庫進行分庫備份，用腳本實現

爲何要進行分庫備份呢，由於，若是在之後須要備份一個小庫，就會很麻煩

常規方法：

mysqldump -B userinfo click test | gzip >bak.sql.gz

分庫備份：

mysqldump -B userinfo | gzip >bak.sql.gz
mysqldump -B click | gzip >bak.sql.gz
mysqldump -B test | gzip >bak.sql.gz

執行命令獲取庫名

mysql -uroot -ppizza123 -e "show databases"   |  grep -v _schema | sed 1d

把密碼放到配置文件中

cat  /etc/my.cnf
[client]
user = root
passwd = pizza123

作完這一步就不用在腳本中添加-u和-p參數

備份腳本

path=/backup
mysql="mysql -uroot -ppizza123"
mysqldump="mysqldump -uroot -ppizza123"
[ -d $path ] || mkdir $path
for dbname in `$mysql -e "show databases;" 2>/dev/null|grep -v _schema | sed 1d`
do
    # 這個命令還有不少參數沒有寫
    $mysqldump -B $dbname |gzip >/backup/${dbname}_$(date +%F).sql.gz
done

六、MySQL分庫、分表備份

常規備份

mysqldump  pizza  test  test1 | gzip > bak.sql.gz

pizza是庫名、test和test1是表名

分庫、分表備份

mysqldump  -B pizza  | gzip >bak.sql.gz
mysqldump pizza test1
mysqldump pizza test2
mysqldump pizza test3

腳本編碼

path=/backup
mysql="mysql -uroot -ppizza123"
mysqldump="mysqldump -uroot -ppizza123"
[ -d $path ] || mkdir $path
for tname in `$mysql -e "show tables from $dbname;" 2>/dev/null|sed 1d`
    do
        if [ "$dbname" = "mysql" ]
        then
            # 這個命令還有不少參數沒有寫
            $mysqldump --skip-lock-tables $dbname $tname |gzip >$path/${dbname}-$tname_$(date +%F).sql.gz 2>/dev/null
        else
            $mysqldump $dbname $tname |gzip >$path/${dbname}-$tname_$(date +%F).sql.gz 2>/dev/null
        fi
    done
done

七、SSH服務批量分發與管理服務

確保主機能用root登錄

vim /etc/ssh/sshd_config 字段 PermitRootLogin 爲 yes

創建密鑰對

ssh-keygen

發送公鑰到其餘服務器

ssh-copy-id  -i  id_rsa.pub  10.0.0.8

可能會很慢，調整其餘機器的配置，讓操做快一些

useDNS no

GSSAPIAuthentication no

重啓服務，繼續

ssh-copy-id  -i  id_rsa.pub  10.0.0.8

ssh-copy-id  -i  id_rsa.pub  10.0.0.9

寫一個連接主機並能夠執行命令的腳本

vim ssh_.sh

if [ $# -ne 1 ]
then
　　echo "usage:$0  cmd"
　　exit 1 
fi
for  n  in 8 9
do
　　echo  "-----10.0.0.$n------"
　　ssh 10.0.0.$n  $1
done

執行腳本

bash  ssh_.sh  "free -m"

編寫分發腳本

. /etc/init.d/functions
if [ $# -ne 2]
then
    echo "usage:$0 localdir remotedir"
    exit 1
fi

for n in 8 9
do
    scp -rp $1 10.0.0.$n:$2 &>/dev/null
    if [ $? -eq 0 ]
    then
        action "10.0.0.$n sucessful" /bin/true
    else
        actin "10.0.0.$n fail" /bin/false
    fi
done

八、破解RANDOM隨機數案例

已知下面這些字符串是經過RANDOM隨機變量 md5sum 後，再截取一部分連續字符串的結果，親個破解這些字符串對用的使用md5sum 處理前的RANDOM對應的數字

21023299

00205d1c

a3da1677

1f6d12dd

890684b

解答：

一、分析

RANDOM的隨機範圍是0-32767 。

顯現須要把範圍內的數字都加密，輸出到md5.log中

二、比較

grep 「890684b」 md5.log | wc -l

三、編碼實現

array=(
21023299
00205d1c
a3da1677
1f6d12dd
890684b
)
md5(){
    for n in {0..32767}
    do
        echo -e "$n\t`echo $n|md5sum`" >> /pizza/md5.log
    done

}
crack_num(){
    for num in ${array[*]}
    do
        find=`grep $num /pizza/md5.log`
        if [ `echo $find|wc -l` -eq 1 ]
        then
            echo $find
        fi
    done
}
main(){
    md5
    crack_num
}
main

第二種方法：egrep實現

[root@web-01 /server/scripts]# array=(
> 21023299
> 00205d1c
> a3da1677
> 1f6d12dd
> 890684b
> )

[root@web-01 /server/scripts]# cmd=`echo ${array[*]}|tr " " "|"`

[root@web-01 /server/scripts]# egrep "$cmd" /pizza/md5.log

修改初版

array=(
21023299
00205d1c
a3da1677
1f6d12dd
890684b
)
md5(){
    for n in {0..32767}
    do
        echo -e "$n\t`echo $n|md5sum`" > /pizza/md5.log &
    done

}
crack_num(){
    cmd=`echo ${array[*]}|tr " " "|"`
    egrep "$cmd" /pizza/md5.log
}

main(){
    md5
    crack_num
}
main

利用time命令對比兩個版本的時間

time sh 8_random_crack.sh

九、檢查多個網站地址是否正常

要求

一、使用shell數組方法，檢測策略精良模擬用戶訪問

二、每10秒種作一次全部的檢測，沒法訪問的輸出報警

三、待檢測網址以下

https://www.cnblogs.com/yxiaodao/

https://www.baidu.com/

檢測工具：

url

curl

wget

腳本編碼

. /etc/init.d/functions
url=(
https://www.cnblogs.com/yxiaodao/
https://www.baidu.com/
)
check_url(){
    wget -t 2 -T 5 -o /dev/null -q $1
    if [ $? -eq 0 ]
    then
        action "$1 is ok" /bin/true
    else
        action "$1 is lost" /bin/false
    fi
}

DealUrl(){
    for url in ${url[*]}
    do
        check_url $url
    done
}

main(){
    while true
    do
        DealUrl
        sleep 10
    done

}
main

修改題目，不用數組，將網址放在文件中，作以下修改

DealUrl(){
    while read line
    do
        check_url $line                                                                                                
    done < ./pizza/url.log
}

有一個問題，在咱們操做完後，會產生大量的網頁文件，由於咱們的命令將網頁下載了

須要在命令中添加參數 -- spider

十、利用Shell編程分析Web日誌解決Dos攻擊生產案例

DOS Deny of Service

DDOS 分佈式dos攻擊

請根據web日誌或者網絡鏈接數，監控當某個IP併發鏈接數或者短期內PV達到100（根據實際狀況設定），即調用防火牆命令封掉對應的IP。

防火牆命令：iptables -l INPUT -s IP地址 -j DROP

分析：

一、web日誌或者網絡鏈接數

　　日誌文件，netstat -an | grep -i est，排序去重

二、判斷PV 或者鏈接數大於100 ，取出IP ，封IP

IP 在日誌的第一列，取到IP---->排序---->統計數量----> 按數量從大到小排序

[root@web-01 /server/scripts]# awk '{print $1}' access_2010-12-8.log |sort|uniq -c|sort -rn
     35 59.33.26.105
     23 123.122.65.226
      8 124.115.4.18

也能經過awk的數組來完成

[root@web-01 /server/scripts]# awk '{S[$1]++}END{for(key in S) print S[key],key}' access_2010-12-8.log |sort -rn
35 59.33.26.105
23 123.122.65.226
8 124.115.4.18

編碼腳本

  9 awk '{S[$1]++}END{for(key in S) print S[key],key}' access_2010-12-8.log |sort -rn > /pizza/ip.log
 10 while read line
 11 do
 12     ip=`echo $line|awk '{print $2}'`
 13     count=`echo $line|awk '{print $1}'`                                                                            
 14     if [ $count -gt 30 -a `grep $ip /pizza/drop.log|wc -l` -lt 1  ]
 15     then
 16         iptables -I INPUT -s $ip -j DROP &&\
 17             echo "$ip" >>/pizza/drop.log
 18     else
 19         echo "$ip" >>/pizza/accept.log
 20     fi
 21 done</pizza/ip.log

本次採用了讀取drop.log日誌的方法，也能夠採用查看 iptables -nL的方法

十、利用Shell編程分析Linux服務器網絡連接數解決DOS攻擊生產案例實踐

仍是上一個題，上面的題監控的是web日誌，本次是監控網絡連接數實現

命令：ESTABLISHED 正在創建的連接狀態

[root@web-01 /server/scripts]# netstat -an |grep -i ESTABLISHED
Active Internet connections (servers and established)
tcp        0      0 172.17.214.84:47778     107.175.240.135:2222    ESTABLISHED
tcp        0      0 172.17.214.84:34820     100.100.30.25:80        ESTABLISHED
tcp        0     52 172.17.214.84:22        163.125.30.51:37793     ESTABLISHED
Active UNIX domain sockets (servers and established)

獲取外部地址，統計，排序（爲方便，將命令的輸出到了日誌）

[root@web-01 ~]# awk '/ESTAB/{print $0}' netstat.log|awk -F "[ :]+" '{print $(NF-3)}'|sort|uniq -c|sort -rn

高級寫法，經過awk數組

[root@web-01 ~]# awk -F "[ :]+" '/ESTAB/{S[$(NF-3)]++}END{for(k in S) print S[k],k}' netstat.log | sort -rn |head

不用日誌，用netstat -an

[root@web-01 ~]# netstat -an|awk -F "[ :]+" '/ESTAB/{S[$(NF-2)]++}END{for(k in S) print S[k],k}' | head
1 163.125.30.51
1 100.100.30.25
1 107.175.240.135
因數據差別，具體命令中參數還要本身調

腳本編寫，只需修改前一個腳本的第一行獲取ip的命令便可

netstat -an|awk -F "[ :]+" '/ESTAB/{S[$(NF-2)]++}END{for(k in S) print S[k],k}'|head >/pizza/ip.log

while read line 
do
    ip=`echo $line|awk '{print $2}'`
    count=`echo $line|awk '{print $1}'`
    if [ $count -gt 30 -a `grep $ip /pizza/drop.log|wc -l` -lt 1  ]
    then
        iptables -I INPUT -s $ip -j DROP &&\
            echo "$ip" >>/pizza/drop.log
    else
        echo "$ip" >>/pizza/accept.log
    fi
done</pizza/ip.log 
在實際工做中，能夠設置定時任務，每3分鐘執行一次，天天晚上0點取消

十一、開發MySQL服務啓動中止腳本

要求：用函數，case語句，if語句等實現 /etc/init.d/mysqld {start | stop | restart} 命令

分析：

一、啓動

mysqld_safe --user=mysql &

二、中止

mysqladmin -uroot -ppasswd shutdown

killall，pkill（參考以前寫的rsync 第九章-case結構條件句）

三、腳本編碼

看一下mysql的pid的文件位置

# 定義鎖文件
lockfile=/var/lock/subsys/mysqld
# 定義變量，指定mysqld的的pid,是須要本身mysql的conf中去建立
mysql_pid_file_path=/application/mysql/data/`uname -n.pid`
. /etc/init.d/functions
start(){
    mysql_safe --user=mysql &>/dev/null &
    retval=$?
    if [ $retval -eq 0 ]
    then
        action  "mysql startup ok" /bin/true
        touch $lockfile
        return $retval
    else
        action "mysql startup fail" /bin/false
        return $retval
    fi
}
stop(){
    if test -s "$mysql_pid_file_path"
    then
        mysql_pid=`cat $mysql_pid_file_path`
        if (kill -0 $mysql_pid &>/dev/null)
        then
            # 爲了在重複中止操做的時候，不提示，將其扔到黑洞
            kill $mysql_pi
            retval=$?
            if [ $? -eq 0 ]
            then
                action "mysql stop ok" /bin/true
                rm -f $lockfile
                return $retval
            else
                action "mysql stop fail" /bin/false
                return $retval
            fi
        else
            echo "mysqld_process is not exit."
            return 2
        fi
    else
        echo "$mysqld_pid_file_path is not exist,or mysqld does not startup"
    fi

}
restart(){
    killall mysql && sleep 1 && mysql --deamon
    retval=$?
    if [ $? -eq 0 ]
    then
        action "mysql restart ok" /bin/true
        return $retval
    else
        action "mysql restart fail" /bin/false
        return $retval
    fi
    }
case "$1" in
    start)
        start
        # 我了向外傳值
        retval=$?
        ;;
    stop)
        stop
        retval=$?
        ;;
    restart)
        stop
        sleep 2
        start
        retval=$?
        ;;

    *)
        echo "usage:$0 {start|stop|restart}"
        exit 1
esac
exit $retval

出現問題，啓動了，可是沒有pid文件

解決：

一、先使用系統的命令開啓。

二、經過查看ps -ef |grep mysql 查看啓動參數

添加啓動參數 --pid-file=$mysql_pid_file_path

問題：進不去mysql

添加參數--datedir=/application/mysql/data

問題：啓動的過程很快，執行腳本後，啓動成功，可是沒有發現進程和pid，沒法進入

一、查看mysql日誌

cat /application/mysql/data/web01.err

二、發現使用腳本中的命令，手動也起不來

三、使用系統執行後的啓動命令

/bin/sh /application/masql/bin/mysqld_safe --datedir=/application/mysql/data --pid-file=$mysql_pid_file_path

腳本必定要如今命令行測試成功，再寫入腳本中

最後一個任務

拷貝到/etc/init.d中，變成chkconfig 可已使用的腳本

十二、按單詞去重排序

In the world of hackers, the kind of answers you get to your technical questions depends as much on the way you ask the questions as on the difficulty of developing the answer. This guide will teach you how to ask questions in a way more likely to get you a satisfactory answer.
Now that use of open source has become widespread, you can often get as good answers from other, more experienced users as from hackers. This is a Good Thing; users tend to be just a little bit more tolerant of the kind of failures newbies often have. Still, treating experienced users like hackers in the ways we recommend here will generally be the most effective way to get useful answers out of them, too.
The first thing to understand is that hackers actually like hard problems and good, thought-provoking questions about them. If we didn't, we wouldn't be here. If you give us an interesting question to chew on we'll be grateful to you; good questions are a stimulus and a gift. Good questions help us develop our understanding, and often reveal problems we might not have noticed or thought about otherwise. Among hackers, 「Good question!」 is a strong and sincere compliment.
Despite this, hackers have a reputation for meeting simple questions with what looks like hostility or arrogance. It sometimes looks like we're reflexively rude to newbies and the ignorant. But this isn't really true.
What we are, unapologetically, is hostile to people who seem to be unwilling to think or to do their own homework before asking questions. People like that are time sinks — they take without giving back, and they waste time we could have spent on another question more interesting and another person more worthy of an answer. We call people like this 「losers」 (and for historical reasons we sometimes spell it 「lusers」).
We realize that there are many people who just want to use the software we write, and who have no interest in learning technical details. For most people, a computer is merely a tool, a means to an end; they have more important things to do and lives to live. We acknowledge that, and don't expect everyone to take an interest in the technical matters that fascinate us. Nevertheless, our style of answering questions is tuned for people who do take such an interest and are willing to be active participants in problem-solving. That's not going to change. Nor should it; if it did, we would become less effective at the things we do best.
We're (largely) volunteers. We take time out of busy lives to answer questions, and at times we're overwhelmed with them. So we filter ruthlessly. In particular, we throw away questions from people who appear to be losers in order to spend our question-answering time more efficiently, on winners.
If you find this attitude obnoxious, condescending, or arrogant, check your assumptions. We're not asking you to genuflect to us — in fact, most of us would love nothing more than to deal with you as an equal and welcome you into our culture, if you put in the effort required to make that possible. But it's simply not efficient for us to try to help people who are not willing to help themselves. It's OK to be ignorant; it's not OK to play stupid.
So, while it isn't necessary to already be technically competent to get attention from us, it is necessary to demonstrate the kind of attitude that leads to competence — alert, thoughtful, observant, willing to be an active partner in developing a solution. If you can't live with this sort of discrimination, we suggest you pay somebody for a commercial support contract instead of asking hackers to personally donate help to you.
If you decide to come to us for help, you don't want to be one of the losers. You don't want to seem like one, either. The best way to get a rapid and responsive answer is to ask it like a person with smarts, confidence, and clues who just happens to need help on one particular problem.

按單詞出現的頻率降序排序

一、把空格和符號都轉換成空格--排序--統計--排序

二、命令

[root@web-01 /server/scripts]# cat english.txt |tr "「」! ,.)( " "\n" |sort|uniq -c |sort -rn

方法二：

[root@web-01 /server/scripts]# cat english.txt |tr "「」! ,.)( " "\n" |awk '{S[$1]++}END{for(k in S) print S[k],k}'|sort -rn

方法三：

cat english.txt |xargs -n1

十二、按字母去重排序

按字母出現的頻率降序排序

一、使用 grep -o ‘.’ 匹配任意以後，會挨個輸出或者 grep -o "[^ ]"

grep -o "[^ ,.()]" english.txt |awk '{S[$1]++}END{for(k in S) print S[k],k}'|sort -rn

二、awk能夠用空作分隔符

sed 's#[ ,\.\]##g' english.txt|awk -F "" '{for(i=0;i<NF;i++)S[$i]++}END{for(k in S) print S[k],k}' |sort -rn

暫時沒有設計出過濾換換行符

1三、按單詞去重排序高級方案

基於上面的awk統計單詞

awk -F "[ ,.]" '{for(i=1;i<NF;i++)S[$i]++}END{for(k in S) print S[k],k}' english.txt |sort -rn

教學例題講解

word文檔

合格的運維人員必會的腳本列表

1）系統及各種服務的監控腳本，例如：文件、內存、磁盤、端口，URL監控報警等。

2）監控網站目錄下文件是否被篡改，以及站點目錄批量被篡改後如何批量恢復的腳本。

3）各種服務Rsync、Nginx、MySQL等的啓動及中止專業腳本（使用chkconfig管理）。

4）MySQL主從複製監控報警以及自動處理不復制故障的腳本。

5）一鍵配置MySQL多實例、一鍵配置MySQL主從部署腳本。

6）監控HTTP/MySQL/Rsync/NFS/Memcached等服務是否異常的生產腳本。

7）一鍵軟件安裝及優化的腳本，好比LANMP、Linux一鍵優化，一鍵數據庫安裝、優化等。

8）MySQL多實例啓動腳本，分庫、分表自動備份腳本。

9）根據網絡鏈接數以及根據Web日誌PV封IP的腳本。

10）監控網站的PV以及流量，而且對流量信息進行統計的腳本。

11）檢查Web服務器多個URL地址是否異常的腳本，要能夠批量處理且通用。

12）系統的基礎優化一鍵優化的腳本。

13）TCP鏈接狀態及IP統計報警腳本。

14）批量建立用戶並設置隨機8位密碼的腳本

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