-2

I want to check in Bash a Variable, if it contains a IP-Adress, otherwise the Script should exit with an errormessage. For example : The var "10.01.24.10" should pass, the var "hello" not, but the var "hello, my ip is 10.03.24.44" should pass not too. My Idea would to use grep but I dont know how to build the grep command at this point. Any Ideas?

madcat
  • 13
  • 5
  • Welcome to Stack Overflow. [SO is a question and answer page for professional and enthusiast programmers](https://stackoverflow.com/tour). Please add your own code to your question. You are expected to show at least the amount of research you have put into solving this question yourself. – Cyrus Oct 16 '20 at 08:35
  • Thank you. I don't have any ideas to solve the problem, so I can't add code. – madcat Oct 16 '20 at 08:38
  • Then at least tell us what did you search for, and what did you find? – tripleee Oct 16 '20 at 08:42
  • See: [Check for IP validity](https://stackoverflow.com/q/13777387/3776858) – Cyrus Oct 16 '20 at 08:42

1 Answers1

0

You can incorporate the grep command in an if statement as below

ip="192.168.240.2" 


if grep -Eq '^([0-9]{1,3}\.){3}([0-9]{1,3})' <<< "$ip"
then 
  echo "PASS"
else 
  echo "FAIL"
fi

You can also write the conditional statements as:

grep -Eq '^([0-9]{1,3}\.){3}([0-9]{1,3})' <<< "$ip" && echo "PASS" || echo "FAIL"

This uses && to signify success and || to signify failure

tripleee
  • 175,061
  • 34
  • 275
  • 318
Raman Sailopal
  • 12,320
  • 2
  • 11
  • 18
  • `#!/bin/bash ip="192.168.240" if grep -Eq '^([0-9]{1,3}.){3}([0-9]{1,3})' <<< $ip then echo "PASS" else echo "FAIL" fi ~ ` If i run this, the script pass too, can i avoid this? – madcat Oct 16 '20 at 08:51
  • The author of this answer forgot to escape the dot `\.`; I have updated to fix this. – tripleee Oct 16 '20 at 08:56