Email Validation Using Loops

Question

I have a question regarding Email validation with Java. I know that the right way of using it is by regular expression and patterns, but for some reason I have been asked to write it using loop. In other words, I have to verify an email using a for loop. It is easy to reach the requirements with patterns, but when it comes to loops, it's a bit confusing.

The email must have at least one letter before @ and at least one letter between @ and dot .. And at least two letters after the dot.

What I meant is how to write a loop that checks all of these requirements using for loop? Any idea?

Earlier today I have asked the question above ... Now I reached here. The only problem now is that I cannot put a minimum of two characters after the dot.

public class EmailValid {
    public static void main (String[] args) {
        System.out.print("Please Enter Your Email To Validate: ");
        Scanner input = new Scanner (System.in);
        String email  = input.nextLine();
        for (int i = 0 ;  i < email.length(); i ++) {
            if (email.indexOf("@") >= 1 ) {
                if (email.indexOf(".") >= email.indexOf("@")+2)
                    if ((email.indexOf("@")+email.indexOf(".") - email.length()) >= 1 )
                        System.out.print("Email Passed" + (email.length() - (email.indexOf("@")+email.indexOf("."))));
            }
        }
        System.out.print("\n String legnth: "+ (email.length()) + 
                "\n Index of @: " + email.indexOf("@") +
                "\n Index of dot: " + email.indexOf("."));
    }
}

you just described it with your last sentence. If you want to have some help is better if you try to show your best try (code).. or at least what did you tried so far.. — LMG, Jun 08 '14 at 15:20
Sorry my bad. What I meant is how can I write a loop that checks all of these requirements using for loop? — Whizz, Jun 08 '14 at 15:21
Maybe looping through the email `String` and noting when encountering '@' and '.' ? — Serge Ballesta, Jun 08 '14 at 15:24
I understand the concept of looping through the string, but how can I put constraints for how many letters should come before or after '@' and '.'. — Whizz, Jun 08 '14 at 15:26
The last constraint is two letters after **the** dot. Your constraint is really only one dot after '@' ??? Or two letters after **last** dot ? — Serge Ballesta, Jun 08 '14 at 15:47
@Whizz- Kindly check the answer and reply if any error found. I hope this method is fine! — Am_I_Helpful, Jun 08 '14 at 15:47
@SergeBallesta at least one letter before '@' and at least one letter before '.' and at least two after the '.'. For example, [email protected] — Whizz, Jun 08 '14 at 16:04
@JarrodRoberson This question is not repeated. The question you are referring to is using patterns. My question requires no patters whatsoever. Please read the question carefully! — Whizz, Jun 08 '14 at 19:46
Please don't edit your question if it changes what you are asking. — Veedrac, Jun 08 '14 at 20:09

Am_I_Helpful · Accepted Answer · 2014-06-08T17:25:04.203

0

Better you should stick to the method which you have adopted. You should traverse the entire String using the for-loop. I think the code must be like this :-

public class EmailValid {
 public static void main (String[] args) {
 System.out.print("Please Enter Your Email To Validate: ");
 Scanner input = new Scanner (System.in);
 String email  = input.nextLine();
 boolean flag=false;
 int countr=0,countd=0;
 loop:for (int i=0; i < email.length();i++) {                 //Better to iterate loop till end of the String "email"
 if(email.charAt(i)=='@'){
  countr++;
  if(countr>1){
   flag=false;
   break loop;
  }
   if(i>=1) 
   flag=true;
   else{
   flag=false;
   break loop;
    }
  }
 if(email.charAt(i)=='.'){
  countd++;
  if(countd>1){
      flag=false;
      break loop;}
   if(i>=3)
 flag=true;
 else{
   flag=false;
   break loop;}
if(email.length()>=i+3)
  flag=true;
 else{
  flag=false;
  break loop;}
 }
 if((email.indexOf(".")-email.indexOf("@"))>=2)
{
  flag=true;}
 else {
  flag=false;
  break loop;}
if(((int)email.charAt(i))>=65 && ((int)email.charAt(i))<=90)
 flag=true;
 else if(((int)email.charAt(i))>=97 && ((int)email.charAt(i))<=122)
 flag=true;
 else if(((int)email.charAt(i))>=48 && ((int)email.charAt(i))<=57)
 flag=true;
 else if(((int)email.charAt(i))==64 || ((int)email.charAt(i))==46)
 flag=true;
 else
 flag=false;
 //@ must at least have a char before itself and also there must be a character between '@' and '.'--So mininmum index of '@' >=1 and for '.' >=3.
 if(flag==false)
break loop;
 }
 if(flag==true && email.length()>=5)    // The minimum length of the String must be equal to minimum index of '.' +2, i.e., >=5. 
  System.out.println("Validated");
  else 
  System.out.println("Invalid Email");
   }
  }

edited Jun 08 '14 at 17:25

answered Jun 08 '14 at 15:46

Am_I_Helpful

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So +=+=+=.+=+=@78 should be a valid address if I have confidence in your code ?!!! – Serge Ballesta Jun 08 '14 at 15:50
You still not verify that '.' comes **after** '@'. IMHO, you should use a state enum ( begin, after_arobas, after_dot), it will be cleaner and more easy to understand. But good tries ... – Serge Ballesta Jun 08 '14 at 16:05
@Whizz-The problem was because of no break option option after becoming false! SORRY. Please try now! – Am_I_Helpful Jun 08 '14 at 16:54
2

@shekharsuman this in now way *it is complete in all aspects like a real "E-Mail"!*, far from it! [ RFC5322](http://tools.ietf.org/html/rfc5322). – Jun 08 '14 at 21:09
If this took 45 mins to get so wrong you should reconsider your profession! – Jun 08 '14 at 21:09
@JarrodRoberson Any suggestions beside his answer? – Whizz Jun 08 '14 at 21:36
@Whizz - yes read the RFC specification for email addresses and understand it before you try to implement something that tests to see if a string is a conformant address. – Jun 09 '14 at 02:56
@JarrodRoberson-Firstly, I meant to say 45 minutes because I was eating while solving his problem. I was doing both work at the same-time. Secondly, my 45 minutes meant only 10+ minutes, this 45 was to make him realise how bad is it to give home-work questions! Lastly, I am a student and still in learning phase; I don't need a change in my profession--My E-mail is in accordance with "Latin characters" allowed for E-mail, not for the Greek,Chinese,etc and neither I care for them. characters-So,I hope replied fine in all aspect. Do you agree OR I am really left with some more points to go on? – Am_I_Helpful Jun 09 '14 at 04:30
@JarrodRoberson-Please check the link below and tell me where it is wrong :- http://tinypic.com/view.php?pic=iojhok&s=8 It's working fine for Latin characters which he and I both wanted!!! Please share your thought. – Am_I_Helpful Jun 09 '14 at 04:44

Email Validation Using Loops

1 Answers1