shell case -14

Write a script to achieve the following functions: 
Enter a number, and then run a corresponding command.

Display command is as follows:

* cmd meau ** 1 - date 2 - ls 3 - who 4 - pwd
When input 1 DATE run, run-time input 2 ls, and so on.

The core elements

• case judgment

  Questions asked

A Ming teacher Answers

Answers

#!/bin/bash
echo "*cmd meau**  1 - date 2 - ls 3 - who 4 - pwd"
read -p "Please input a number: " n
if [ -z "$n" ]
then
    echo "请输入一个纯数字,范围1-4."
    exit
fi

n1=`echo $n|sed 's/[0-9]//g'`
if [ -n "$n1" ]
then
    echo "请输入一个纯数字，范围1-4."
    exit
fi

case $n in 
    1)
    date
    ;;
    2)
    ls
    ;;
    3)
    who
    ;;
    4)
    pwd
    ;;
    *)
    echo "请输入1-4的数字"
        ;;
esac

Questions asked

Shell script with the following requirements:

Add user_00 - user_09 10 users and give them to set a random password, password requirements 10 contains uppercase and lowercase letters and numbers, pay attention to the needs of each user's password records to a log file.
 
prompt:

1. Random password using the command mkpasswd

2. User password in the script, then you can use the echo command passwd pipeline

The core elements

• seq digital increments
• mkpasswd generate random characters

seq -w 00 09 achieve two consecutive numbers
mkpasswd -l 10 // 10 characters
mkpasswd -l 10 -s 0 // 0 and no special symbols special symbols of

echo -e "passwd \ npasswd \ n " | passed user1 method for a user to set a password for the user
echo "sdfewdsfewfew12d" | passwd --stdin user Method Two user to set a password for the user

A Ming teacher Answers

Answers

#!/bin/bash
for i in `seq -w 00 09`
do
    useradd user_$i
    p=`mkpasswd -l 10 -s 0 `
    echo "user_$i $p" >> /tmp/pass.tmp
    echo $p |passwd --stdin user_$i
done

Questions asked

On the server, write a script monitoring requirements are as follows:

1. Every 10s to detect the number of httpd processes on a server, if 500 or more of the time, you need to automatically restart the apache service, and to detect the success of start?

2. If you do not start normally need to start again, more than five times the maximum number of unsuccessful need to immediately notify the administrator e-mail, and later do not need to detect!

3. If successful start, one minute after the test again a few httpd process, if repeated before the normal operation (tested once every 10s), if greater than or equal to 500, reboot and give up that need to send e-mail to the administrator, and then automatically exit the script . Assume mail.py mail script which is used before

The core elements

• pgrep -l httpd or ps -C httpd --no-heading the inspection process
• for 5 cycles counter

A Ming teacher Answers

Answers

#!/bin/bash
check_service()
{
    n=0
    for i in `seq 1 5`
    do
        /usr/local/apache2/bin/apachectl restart 2>/tmp/apache.err
        if [  ！$? -ne 0 ]
        then
            n=$[$n+1]
        else 
            break
        fi
    done
    if [ $n -eq 5 ]
    then
        ##下面的mail.py参考https://coding.net/u/aminglinux/p/aminglinux-book/git/blob/master/D22Z/mail.py
        python mai.py "[email protected]" "httpd service down" `cat /tmp/apache.err`
        exit
    fi
}   
while true
do
    t_n=`ps -C httpd --no-heading |wc -l`
    if [ $t_n -ge 500 ]
    then
        /usr/local/apache2/bin/apachectl restart
        if [ $? -ne 0 ]
        then
            check_service
        fi
        sleep 60
        t_n=`ps -C httpd --no-heading |wc -l`
        if [ $t_n -ge 500 ]
        then
            python mai.py "[email protected]" "httpd service somth wrong" "the httpd process is busy."
            exit
        fi
    fi
    sleep 10
done

#### 题目要求
需求： 根据web服务器上的访问日志，把一些请求量非常高的ip给拒绝掉！并且每隔半小时把不再发起请求或者请求量很小的ip给解封。
 
假设： 

1. 一分钟内请求量高于100次的IP视为不正常请求。

2. 访问日志路径为/data/logs/access_log。

用第2例中的1.log作为演示日志

#### 核心要点
* 统计ip访问次数，排序
* 如何标记每隔半小时
* iptables计数器是一个重要的判断指标
* 函数（封IP、解封IP）

有点难需要好好想想 
#### 参考答案

#!/bin/bash
block_ip()
{
t1=date -d "-1 min" +%Y:%H:%M
log=/data/logs/access_log

egrep "$t1:[0-9]+" $log > /tmp/tmp_last_min.log
awk '{print $1}' /tmp/tmp_last_min.log |sort -n |uniq -c|sort -n |awk '$1>100 {print $2}' > /tmp/bad_ip.list
n=wc -l /tmp/bad_ip.list|awk '{print $1}'
if [ $n -ne 0 ]
then
for ip in cat /tmp/bad_ip.list
do
iptables -I INPUT -s $ip -j REJECT
done
fi
}

unblock_ip()
{
iptables -nvL INPUT|sed '1d' |awk '$1<5 {print $8}' > /tmp/good_ip.list
n=wc -l /tmp/good_ip.list|awk '{print $1}'
if [ $n -ne 0 ]
then
for ip in cat /tmp/good_ip.list
do
iptables -D INPUT -s $ip -j REJECT
done
fi
iptables -Z
}

t=date +%M
if [ $t == "00" ] || [ $t == "30" ]
then
unblock_ip
block_ip
else
block_ip
fi

#### 题目要求
请仔细查看如下几个数字的规律，并使用shell脚本输出后面的十个数字。

10 31 53 77  105 141 …….

#### 核心要点
* 计算两个数值之间的差值

    21   22   24    28   36  

#### 参考答案

#!/bin/bash
x=10
y=21
for i in seq 0 15
do
echo $x
x=$[$x+$y]
z=$[2**$i]
y=$[$y+$z]
done