如果我可能没有收到所有4个八位位组作为输入，如何使用Bash将IP地址分成4个八位位组变量？

Question

I am writing a script that will be running in bash.

The expected input for this program will look like one of the following 4 cases:

run 192.168.1.1 [Other Parameters]

run 168.1.1 [Other Parameters]

run 1.1 [Other Parameters]

run 1 [Other Parameters]

Given any of these cases I want to have the following array.

IP[0]=192

IP[1]=168

IP[2]=1

IP[3]=1

The 4th Octet will always be required. If the others are not supplied they will default to 192.168.1.x.

I saw this page on seperating Octets but I was not sure how to account for the 4 different input cases I will need to handle. Here.

Any help or suggestions on how I could improve or implement this would be greatly appreciated.

Thanks.

Answer 1

First, split the IP address into 4 parts, but only after "padding" it to guarantee 4 fields:

input=168.1.1
IFS=. read -a o <<< "...$input"

Then, build the patched output, using default-value parameter expansion to patch the fields that were assigned null strings from the padding.

declare -a ip=( ${o[-4]:-192}
                ${o[-3]:-168}
                ${o[-2]:-1}
                ${o[-1]:-1} )

And finally, you can stitch it back together:

output=$( IFS=.; echo "${ip[*]}" )

Answer 2

This should work. It uses a brute force to find the length of the IP, and the $IFS variable to separate it into the array:

#! /bin/bash
function full_ip () {
    if [[ $1 = *.*.*.* ]] ; then
        echo $1
    elif [[ $1 = *.*.* ]] ; then
        echo 192.$1
    elif [[ $1 = *.* ]] ; then
        echo 192.168.$1
    else
        echo 192.168.1.$1
    fi

}

for ip in 192.168.1.1 168.1.1 1.1 1 ; do
    f=$(full_ip $ip)
    echo -n $f$'\t'
    IFS=. IP=($f) IFS=$' \t\n'
    echo ${IP[0]} ${IP[1]} ${IP[2]} ${IP[3]}
done