bash脚本中替换变量的可选参数

Question

How can I pass an optional argument to a bash script that will replace an existing variable in the script? For example:

#!/bin/bash
#hostfinder.sh
#Find hosts in current /24 network

subnet=$(hostname -i | cut -d. -f1,2,3)

for j in \
    $(for i in $subnet.{1..255}; do host $i; done | grep -v not | cut -d" " -f5)
do ping -c1 -w 1 $j; done | grep -i from | cut -d" " -f3,4,5 | tr ':' ' ' | \
sed -e 's/from/Alive:/'

This will grab the IP of the current host, run a reverse lookup against possible neighbors, ping test any hostnames it finds, and display an output similar to this:

Alive: host1.domain (10.64.17.23)
Alive: host2.domain (10.64.17.24)
...

Call me crazy, but it's way faster than nmap and spits out a nice list.

Anyway, I'd like to optionally pass the first three octets of any class C network address to the $subnet variable as the $1 argument when executing the script. For example:

./hostfinder.sh 10.20.0

My first thought was to try something like $subnet=$1, but I assume that will not work. I'm not really interested in re-writing the script to be more elegant or anything, I am mostly just curious about what I put in the subject line.

Answer 1

How about replace:

subnet=$(hostname -i | cut -d. -f1,2,3)

with:

case $# in  
  0) subnet=$(hostname -i | cut -d. -f1,2,3);;
  1) subnet="${1}";;
  *) echo "To many arguments" >&2; exit 1;;
esac

• $# is the number of command line arguments
• This is less elegant as getopt but easy to understand and extend.

Answer 2

尝试使用getopt从命令行中的选项中读取，然后设置变量。

Answer 3

Like LtWorf suggested, try using getopt . Its read the options and its arguments from the command line.

You can find a great example of getopt usage here: getopt usage example