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++

// 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

Java

// 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");
    }
}

Python3

# 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:
url = "https://www.geeksforgeeks.org"
 
if(isValidURL(url) == True):
    print("Yes")
else:
    print("No")
 
# This code is contributed by avanitrachhadiya2155

Output:

Yes

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

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