How to validate an IP address using Regular Expressions in Java

10 Jun, 2021
Given an IP address, the task is to validate this IP address with the help of Regular Expressions.The IP address is a string in the form “A.B.C.D”, where the value of A, B, C, and D may range from 0 to 255. Leading zeros are allowed. The length of A, B, C, or D can’t be greater than 3.Examples:  
Input: str = "000.12.12.034"
Output: True

Input: str = "000.12.234.23.23"
Output: False

Input: str = "121.234.12.12"
Output: True

Input: str = "I.Am.not.an.ip"
Output: False
 
Approach: In this article, the IP address is validated with the help of Regular Expressions or Regex. Below are the steps to solve this problem using ReGex: 
Get the string.Regular expression to validate an IP address:  
Get the string.
Regular expression to validate an IP address:  
// ReGex to numbers from 0 to 255
zeroTo255 -> (\\d{1, 2}|(0|1)\\d{2}|2[0-4]\\d|25[0-5])

// ReGex to validate complete IP address
IPAddress -> zeroTo255 + "\\." + zeroTo255 
                + "\\." + zeroTo255 
                + "\\." + zeroTo255;
where: \d represents digits in regular expressions, same as [0-9]\\d{1, 2} catches any one or two-digit number(0|1)\\d{2} catches any three-digit number starting with 0 or 1.2[0-4]\\d catches numbers between 200 and 249.25[0-5] catches numbers between 250 and 255.Match the string with the Regex. In Java, this can be done using Pattern.matcher().  Return true if the string matches with the given regex, else return false.
where: \d represents digits in regular expressions, same as [0-9]\\d{1, 2} catches any one or two-digit number(0|1)\\d{2} catches any three-digit number starting with 0 or 1.2[0-4]\\d catches numbers between 200 and 249.25[0-5] catches numbers between 250 and 255.
\d represents digits in regular expressions, same as [0-9]
\\d{1, 2} catches any one or two-digit number
(0|1)\\d{2} catches any three-digit number starting with 0 or 1.
2[0-4]\\d catches numbers between 200 and 249.
25[0-5] catches numbers between 250 and 255.
Match the string with the Regex. In Java, this can be done using Pattern.matcher().  
Return true if the string matches with the given regex, else return false.
Below is the implementation of the above approach: 
Java
// Java program to validate an IP address// using Regular Expression or ReGex import java.util.regex.*; class IPAddressValidation {     // Function to validate the IPs address.    public static boolean isValidIPAddress(String ip)    {         // Regex for digit from 0 to 255.        String zeroTo255            = "(\\d{1,2}|(0|1)\\"              + "d{2}|2[0-4]\\d|25[0-5])";         // Regex for a digit from 0 to 255 and        // followed by a dot, repeat 4 times.        // this is the regex to validate an IP address.        String regex            = zeroTo255 + "\\."              + zeroTo255 + "\\."              + zeroTo255 + "\\."              + zeroTo255;         // Compile the ReGex        Pattern p = Pattern.compile(regex);         // If the IP address is empty        // return false        if (ip == null) {            return false;        }         // Pattern class contains matcher() method        // to find matching between given IP address        // and regular expression.        Matcher m = p.matcher(ip);         // Return if the IP address        // matched the ReGex        return m.matches();    }     // Driver code    public static void main(String args[])    {        // Checking for True case.        // Test Case: 1        System.out.println("Test Case 1:");        String str1 = "000.12.12.034";        System.out.println("Input: " + str1);        System.out.println(            "Output: "            + isValidIPAddress(str1));         // Test Case: 2        System.out.println("\nTest Case 2:");        String str2 = "121.234.12.12";        System.out.println("Input: " + str2);        System.out.println(            "Output: "            + isValidIPAddress(str2));         // Checking for False case.        // Test Case: 3        System.out.println("\nTest Case 3:");        String str3 = "000.12.234.23.23";        System.out.println("Input: " + str3);        System.out.println(            "Output: "            + isValidIPAddress(str3));         // Test Case: 4        System.out.println("\nTest Case 4:");        String str4 = "I.Am.not.an.ip";        System.out.println("Input: " + str4);        System.out.println(            "Output: "            + isValidIPAddress(str4));    }}
Test Case 1:
Input: 000.12.12.034
Output: true

Test Case 2:
Input: 121.234.12.12
Output: true

Test Case 3:
Input: 000.12.234.23.23
Output: false

Test Case 4:
Input: I.Am.not.an.ip
Output: false
 
clintra
java-regular-expression
Technical Scripter 2019
Computer Networks
Java
Technical Scripter
Java
Computer Networks
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