How would I run an ifconfig, grab the IPV4 address, place it in a variable and then output the results in shell?

Question

I hope all is well,

I'm struggling to put something together so I thought I would ask here.

I'm trying to find a way to right a shell script that

• Runs ifconfig
• grabs just the inet (IPV4)
• Place it in a variable
• Echo "My IP"
• Then print the variables value

I was able to do this successfully in bash with both a ping command:

@for /f "delims=[] tokens=2" %%a in ('ping -4 -n 1 %ComputerName% ^|
findstr [') do (
    set "MY_IP=%%a"
)
)

echo IP of Current Computer : %MY_IP%
pause>nul & exit

I'm surprised how much I'm struggling with this. Even a solid resource would help me a lot.

Thank you for your time.

Kind regards,

Answer 1

First of all, if you're looking for your public ip address, you can simply use

ip=$(hostname -I)
echo $ip

Also if you insist on using ifconfig command, you must specify your interface, if your interface for example is 'eth0': (first command from this)

ip=$(ifconfig eth0 | grep 'inet addr' | cut -d: -f2 | awk '{print $1}')
echo $ip

Answer 2

use one of the following

    #!/usr/bin/bash
    IP=`ifconfig wlp3s0 | grep -w inet | cut -d" " -f10 `
    echo $IP

    IP2=`ip addr show wlp3s0 | grep -Po 'inet \K[\d.]+'`
    echo $IP2

replace wlp3s0 with your device name. I would suggest writing your code on "ip addr" instead of ifconfig, as ifconfig may be deprecated on some machines.

I came up with this script which figures all the interfaces and then prints only the ones which have an IP address assigned

    #!/usr/bin/bash
    
    readarray -t interfaces < <(ip l | awk -F ":" '/^[0-9]+:/{dev=$2 ; if ( dev !~ /^ lo$/) {print $2}}')
    for i in "${interfaces[@]// /}" ; 
        do 
            IP=`ip addr show $i | grep -Po 'inet \K[\d.]+'`
            if [ -n $IP ]; then
                echo "Interface is $i"
                echo $IP
            fi
        done

BTW, the above script ignores localhost (lo)

Answer 3

I found that this can do the job

address=$(ifconfig | grep -Eo '[0-9]+\.[0-9]+\.[0-9]+\.[0-9]+' | grep -Ev '(255|127)')
echo ${address}

I just used a regular expression to grab only the ip format and then exclude localhost and the gateway

Answer 4

As you have not clarified whether you're looking for a windows batch-file solution, or an ubuntu linux bash shell script, and those 'nix based tags have now been edited out, the following is an example of a Windows batch-file, which should achieve what your provided code attempted to do. Use the ping command against the computer name to return the IPv4 address, save it as a variable and print the details to the host.

I have written it in such a way as to allow it to function even in Windows 2000, as earlier ping.exe versions, did not have the -4 option.

@Echo Off
SetLocal EnableExtensions
Set "MY_IP="
For /F "Tokens=2 Delims=[]" %%G In (
    '"("%SYSTEMROOT%\System32\ping.exe" "%COMPUTERNAME%" -n 1 -4 2>NUL || "%SYSTEMROOT%\System32\ping.exe" "%COMPUTERNAME%" -n 1 2>NUL) | "%SYSTEMROOT%\System32\find.exe" /I "%COMPUTERNAME%""'
) Do Set "MY_IP=%%G"
If Not Defined MY_IP GoTo :EOF
Echo IP of Current Computer : %MY_IP%
Pause

If you intend using it only on modern versions of Windows OS, then the following shorter version should suffice:

@Echo Off
SetLocal EnableExtensions
Set "MY_IP="
For /F "Tokens=2 Delims=[]" %%G In (
    '""%__APPDIR__%ping.exe" "%COMPUTERNAME%" -n 1 -4 2>NUL | "%__APPDIR__%find.exe" /I "%COMPUTERNAME%""'
) Do Set "MY_IP=%%G"
If Not Defined MY_IP GoTo :EOF
Echo IP of Current Computer : %MY_IP%
Pause