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Check if an URL is valid or not using Regular Expression
  • Last Updated : 11 Feb, 2021

Given a URL as a character string str of size N.The task is to check if the given URL is valid or not.
Examples : 

Input : str = “https://www.geeksforgeeks.org/” 
Output : Yes 
Explanation : 
The above URL is a valid URL.
Input : str = “https:// www.geeksforgeeks.org/” 
Output : No 
Explanation : 
Note that there is a space after https://, hence the URL is invalid. 
 

Approach : 
An approach using java.net.url class to validate a URL is discussed in the previous post. 
Here the idea is to use Regular Expression to validate a URL. 

  • Get the URL.
  • Create a regular expression to check the valid URL as mentioned below:

regex = “((http|https)://)(www.)?” 
+ “[a-zA-Z0-9@:%._\\+~#?&//=]{2,256}\\.[a-z]” 
+ “{2,6}\\b([-a-zA-Z0-9@:%._\\+~#?&//=]*)”
 

  • The URL must start with either http or https and
  • then followed by :// and
  • then it must contain www. and
  • then followed by subdomain of length (2, 256) and
  • last part contains top level domain like .com, .org etc.
  • Match the given URL with the regular expression. In Java, this can be done by using Pattern.matcher().
  • Return true if the URL matches with the given regular expression, else return false.

Below is the implementation of the above approach:
 

C++

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// C++ program to validate URL
// using Regular Expression
#include <iostream>
#include <regex>
using namespace std;
 
// Function to validate URL
// using regular expression
bool isValidURL(string url)
{
 
  // Regex to check valid URL
  const regex pattern("((http|https)://)(www.)?[a-zA-Z0-9@:%._\\+~#?&//=]{2,256}\\.[a-z]{2,6}\\b([-a-zA-Z0-9@:%._\\+~#?&//=]*)");
 
  // If the URL
  // is empty return false
  if (url.empty())
  {
     return false;
  }
 
  // Return true if the URL
  // matched the ReGex
  if(regex_match(url, pattern))
  {
    return true;
  }
  else
  {
    return false;
  }
}
 
// Driver Code
int main()
{
  string url = "https://www.geeksforgeeks.org";
 
  if (isValidURL(url))
  {
    cout << "YES";
  }
  else
  {
    cout << "NO";
  }
  return 0;
}
 
// This code is contributed by yuvraj_chandra

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Java

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// Java program to check URL is valid or not
// using Regular Expression
 
import java.util.regex.*;
 
class GFG {
 
    // Function to validate URL
    // using regular expression
    public static boolean
    isValidURL(String url)
    {
        // Regex to check valid URL
        String regex = "((http|https)://)(www.)?"
              + "[a-zA-Z0-9@:%._\\+~#?&//=]"
              + "{2,256}\\.[a-z]"
              + "{2,6}\\b([-a-zA-Z0-9@:%"
              + "._\\+~#?&//=]*)";
 
        // Compile the ReGex
        Pattern p = Pattern.compile(regex);
 
        // If the string is empty
        // return false
        if (url == null) {
            return false;
        }
 
        // Find match between given string
        // and regular expression
        // using Pattern.matcher()
        Matcher m = p.matcher(url);
 
        // Return if the string
        // matched the ReGex
        return m.matches();
    }
 
    // Driver code
    public static void main(String args[])
    {
        String url
            = "https://www.geeksforgeeks.org";
        if (isValidURL(url) == true) {
            System.out.println("Yes");
        }
        else
            System.out.println("NO");
    }
}

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Python3

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# Python3 program to check
# URL is valid or not
# using regular expression
import re
 
# Function to validate URL
# using regular expression
def isValidURL(str):
 
    # Regex to check valid URL
    regex = ("((http|https)://)(www.)?" +
             "[a-zA-Z0-9@:%._\\+~#?&//=]" +
             "{2,256}\\.[a-z]" +
             "{2,6}\\b([-a-zA-Z0-9@:%" +
             "._\\+~#?&//=]*)")
     
    # Compile the ReGex
    p = re.compile(regex)
 
    # If the string is empty
    # return false
    if (str == None):
        return False
 
    # Return if the string
    # matched the ReGex
    if(re.search(p, str)):
        return True
    else:
        return False
 
# Driver code
 
# Test Case 1:
 
if(isValidURL(url) == True):
    print("Yes")
else:
    print("No")
 
# This code is contributed by avanitrachhadiya2155

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Output: 

Yes

 

Time Complexity: O (N) 
Auxiliary Space: O (1)

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