Email Validation

**juhiswt** · 03-07-2011, 06:08 AM

Java Code:

// Checks for invalid characters
//in email addresses
import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.Scanner;

public class EmailValidation {
   public static void main(String[] args) 
                                 throws Exception {
	  String input=new String();
			char[] id=new char[30];
			char[] server=new char[30];
	Matcher m;
	Pattern p;
 	Scanner S=new Scanner(System.in);
      //Checks for email addresses starting with
      //inappropriate symbols like dots or @ signs.
       do
	{ 
		
		System.out.println("Enter the email Id");
               input = S.nextLine();
		 p = Pattern.compile("^\\.|^\\@");
     		 m = p.matcher(input);
      		if (m.find())
         	System.err.println("Email addresses don't start" +
                            " with dots or @ signs this is not the valid address");
     		 //Checks for email addresses that start with
     		 //www. and prints a message if it does.
      		p = Pattern.compile("^www\\.");
     		 m = p.matcher(input);
      		if (m.find()) 
		{
        		System.out.println("Email addresses don't start" +
                	" with \"www.\", only web pages do,this is not the valid address");
      		}
      		p = Pattern.compile("[^A-Za-z0-9\\.\\@_\\-~#]+");
      		m = p.matcher(input);
	}while(m.find());
	
	System.out.println("It is valid address");
	for(int i=0;i<input.length();i++)
	{
		do{
		 id[i]=input.charAt(i);}while(input.charAt(i)!='@');
		
			for(int j=i;j<input.length();j++)
			server[j]=input.charAt(j );
		
	}
	System.out.println("Email Id:");
	for(int i=0;i<id.length-1;i++)
		System.out.print(id[i]);
	System.out.println("");
	
	System.out.println("Server:");
	for(int i=0;i<server.length-1;i++)
		System.out.print(server[i]);
            
   }
}

for loop is not terminating.I dont know whats the problem last staments are not printing?

**Eranga** · 03-07-2011, 07:53 AM

Java Code:

do {
                id[i] = input.charAt(i);
            } while (input.charAt(i) != '@');

Think about the above code segment. You've not change the value of i in that loop, and there is no way to jump out.

**Eranga** · 03-07-2011, 07:54 AM

And one more thing, please use code tags when you are posting again in the forum. Unformated codes are really hard to read. If you don't know how to do that, check my forum signature. You can find a relevant link there.