Question

Why this code doesn’t raise a ‘ParseException’ ?

DateFormat formatter = new SimpleDateFormat("dd/mm/yyyy");
try {
     String date = "01/01/200'";
     formatter.parse(date);
} catch (ParseException ex) {
     throw new ParseException("Formato de fecha inválido",0);
}

Please explain that to me, I’m lost on that. And note the simple quote '

Answer 1

Because it’s what is supposed to do.
If you check the docs on DateFormat it says about the parameter.

source - A String whose beginning should be parsed.

If you want to restrict the format, you’ll have to use regular expressions.

Also check the rules of SimpleDateformat about the year:

For parsing, if the number of pattern letters is more than 2, the year is interpreted literally, regardless of the number of digits. So using the pattern "MM/dd/yyyy", "01/11/12" parses to Jan 11, 12 A.D.