Date Format (dd/mm/yyyy) Regular Expression Pattern

(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[012])/((19|20)\\d\\d)

The above regular expression is used to validate the date format in “dd/mm/yyyy”, you can easy customize to suit your need. However, it’s a bit hard to validate the leap year, 30 or 31 days of a month, we may need basic logic as below.

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class DateValidator {

  private Pattern pattern;
  private Matcher matcher;

  private static final String DATE_PATTERN =
          "(0?[1-9]|[12][0-9]|3[01])/(0?[1-9]|1[012])/((19|20)\\d\\d)";

  public DateValidator() {
	  pattern = Pattern.compile(DATE_PATTERN);
  }

  /**
   * Validate date format with regular expression
   * @param date date address for validation
   * @return true valid date fromat, false invalid date format
   */
   public boolean validate(final String date) {

     matcher = pattern.matcher(date);

     if(matcher.matches()) {

	 matcher.reset();

	 if(matcher.find()) {

             String day = matcher.group(1);
	     String month = matcher.group(2);
	     int year = Integer.parseInt(matcher.group(3));

	     if (day.equals("31") &&
		  (month.equals("4") || month .equals("6") || month.equals("9") ||
                  month.equals("11") || month.equals("04") || month .equals("06") ||
                  month.equals("09"))) {
			return false; // only 1,3,5,7,8,10,12 has 31 days
	     } else if (month.equals("2") || month.equals("02")) {
                  //leap year
		  if(year % 4==0) {
			  if(day.equals("30") || day.equals("31")) {
				  return false;
			  } else {
				  return true;
			  }
		  } else {
		         if(day.equals("29")||day.equals("30")||day.equals("31")) {
				  return false;
		         } else {
				  return true;
			  }
		  }
	      } else {
		return true;
	      }
	   } else {
    	      return false;
	   }
     } else {
	  return false;
     }
   }
}