What should I change the pattern to validate?

Question

I must validate two types of expressions for an email:

letter.letter.year greater than 2018@uts.edu.co

jg2019@uts.edu.co

0-2018.letter.letter@*.com.co

0.jg@cloudmail.com.co

I have used these two regex expressions but they have not worked:

[a-zA-Z] + [.]? [a-zA-Z] + [.]? [2-9] [0-9] (?! 18 $) [1-9] [1-9] ? + $ \\ @ uts ([.]) edu ([.]) co

\\ b ([0] | 20 [0-1] [0-8] | 2019) \\ b + [.]? [a-zA-Z] + [.]? [a-zA-Z] + \\ @ [ a-zA-Z] ([.]) com ([.]) co

private void btn_validarActionPerformed(java.awt.event.ActionEvent evt) {                                            
String w_correo = caja_correo.getText();
Pattern p_correo1 = Pattern.compile("^[a-zA-Z]+[.]?[a-zA-Z]+[ .]?[2-9][0-9](?!18$)[1-9][1-9]?+$\\@uts([.])edu([\\.])co$");
    Matcher m_correo1 = p_correo1.matcher(w_correo);
Pattern p_correo2 = Pattern.compile("^\\b([0]|20[0-1][0-8]|2019)\\b+[.]?[a-zA-Z]+[.]?[a-zA-Z]+\\@ [a-zA-Z] ([.])com([\\.])co$");
    Matcher m_correo2 = p_correo2.matcher(w_correo);    

    if (m_correo1.matches()) {

String validacion = "";
validacion = validacion +  "Direccion de correo electrónico correcta<br/>";
correcto.setForeground(Color.GREEN);

}

    else { 
String validacion = "";
    if (!m_correo1.matches()) {
    validacion= validacion + "Direccion de correo electrónico incorrecta<br/>";
            incorrecto.setBackground(Color.RED);
    }
}

    if (m_correo2.matches()) { 

String validacion = "";
validacion = validacion +  "Direccion de correo electrónico correcta<br/>";
correcto.setForeground(Color.GREEN);
}

    else {String validacion = "";
    if (!m_correo2.matches()) {
    validacion= validacion + "Direccion de correo electrónico incorrecta<br/>";
            incorrecto.setBackground(Color.RED);
    }
}
}                                           

When you tried to validate a valid email the result is that the email is incorrect. the button NO is shown red, but the button SI is not shown green.

Answer 1

The regex below matches on the pattern you describe: "letter.letter.year greater than 2018@uts.edu.co".

([a-zA-Z]\.){2}([2-9][1-9][0-9]{2}|20([2-9][0-9]|19))@uts\.edu\.co

([a-zA-Z]\.){2} matches any letter followed by a period two times.
([2-9][1-9][0-9]{2}|20([2-9][0-9]|19)) matches the year 2019 or greater
@uts\.edu\.co matches @uts.edu.co literally.

Try it out here: https://regex101.com/r/hNIMRL/1

Answer 2

Try the following regex: "(?:[a-zA-Z]\\.){2}([0-9]{4})\\@uts.edu.co"

I have tested it for your example emails. Using a capture group for the year you can validate that the year is greater than 2018. The RegEx will only match emails that follow your pattern of Letter.Letter.Four Digit Year@uts.edu.co

Here is the .java I used to test the regEx:

import java.util.regex.Pattern;

public class MainApp {

public static void main(String[] args) {

   // reg ex to use:  (?:[a-zA-Z]\.){2}([0-9]{4})\@uts.edu.co

    String[] email = {"j.g.2018@uts.edu.co","s.c.2019@uts.edu.co","t.t.2020@gmail.com"};

    String regExPattern = "(?:[a-zA-Z]\\.){2}([0-9]{4})\\@uts.edu.co";

    Pattern p = Pattern.compile( regExPattern );

    for(String x : email){
            Matcher m = p.matcher( x );
            boolean b = m.matches();
            if(b){
                int year = Integer.parseInt(m.group(1));
                if(year > 2018){
                System.out.println(x + " is valid");
                }
                else{
                System.out.println(x + " is not valid");
                }
            }
            else{
            System.out.println(x + " is not valid");
            }
        }

}}

Log from my console:

jg2018@uts.edu.co is not valid

sc2019@uts.edu.co is valid

tt2020@gmail.com is not valid

Cheers!

Answer 3

For this type of e-mail letter.letter.year greater than 2018@uts.edu.co like jg2019@uts.edu.co you can use:

^[a-zA-Z]\.[a-zA-Z]\.(?:2019|20[2-9][0-9]|2[1-9][0-9]{2}|[3-9][0-9]{3})@uts\.edu\.co$

• ^ Start of string
• [a-zA-Z]\\.[a-zA-Z]\\. Match 2 times a char az followed by a dot
• (?:2019|20[2-9][0-9]|2[1-9][0-9]{2}|[3-9][0-9]{3}) Match range greater than 2018
• @uts\\.edu\\.co Match @uts.edu.co
• $ End of string

In Java

String regex = "^[a-zA-Z]\\.[a-zA-Z]\\.(?:2019|20[2-9][0-9]|2[1-9][0-9]{2}|[3-9][0-9]{3})@uts\\.edu\\.co$";

Regex demo

For this type of e-mail 0-2018.letter.letter@*.com.co like 0.jg@cloudmail.com.co you could use:

^(?:2018|201[0-7]|200[0-9]|1[0-9]{1,3}|[0-9]{1,3})\.[a-zA-Z]\.[a-zA-Z]@\w+(?:\.\w+)*\.com\.co$

• ^ Start of string
• (?:2018|201[0-7]|200[0-9]|1[0-9]{1,3}|[0-9]{1,2}) Range from 0 to 2018
• \\.[a-zA-Z]\\.[a-zA-Z] Match 2 times a dot followed by a char az
• @\\w+(?:\\.\\w+)* Match @, repeat 1+ times a word char followed by repeating 0+ times a dot and 1+ word chars
• \\.com\\.co Match .com.co
• $ End of string

In Java

String regex = "^(?:2018|201[0-7]|200[0-9]|1[0-9]{1,3}|[0-9]{1,3})\\.[a-zA-Z]\\.[a-zA-Z]@\\w+(?:\\.\\w+)*\\.com\\.co$";

Regex demo