Bash脚本用于计算汇总的IP地址范围

Question

我需要编写一个bash脚本,当我输入两个ip地址时,它将为它们计算汇总地址.

Examlpe:

192.168.1.27/25
192.168.1.129/25  

结果将是:

192.168.1.0/24  

你能用这个脚本帮我吗？

我知道你会对我说"你做了什么？"

我试图在谷歌找到一些东西,但我发现我必须转换为二进制然后计算它,这将是非常困难的.

我甚至不知道如何开始.

有什么想法或提示吗？

Answer 1

使用bash计算通用网络掩码:

#!/bin/bash

D2B=({0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1}{0..1})
declare -i c=0                              # set integer attribute

# read and convert IPs to binary
IFS=./ read -r -p "IP 1: " a1 a2 a3 a4     # e.g. 192.168.1.27
b1="${D2B[$a1]}${D2B[$a2]}${D2B[$a3]}${D2B[$a4]}"

IFS=./ read -r -p "IP 2: " a1 a2 a3 a4     # e.g. 192.168.1.129
b2="${D2B[$a1]}${D2B[$a2]}${D2B[$a3]}${D2B[$a4]}"

# find number of same bits ($c) in both IPs from left, use $c as counter
for ((i=0;i<32;i++)); do
  [[ ${b1:$i:1} == "${b2:$i:1}" ]] && c=c+1 || break
done    

# create string with zeros
for ((i=c;i<32;i++)); do
  fill="${fill}0"
done    

# append string with zeros to string with identical bits to fill 32 bit again
new="${b1:0:$c}${fill}"

# convert binary $new to decimal IP with netmask
new="$((2#${new:0:8})).$((2#${new:8:8})).$((2#${new:16:8})).$((2#${new:24:8}))/$c"
echo "$new"

输出:

192.168.1.0/24