Cannot get date RegEx mm/dd/yyyy down in java

Use a try/catch block where you extract the month and day, and then based on the parsed parameters, throw an illegalargument exception. Your situation definitely falls under the circumstances of when to throw such an exception as described here:

When should an IllegalArgumentException be thrown?

Thanks!

Two steps, one less problem¹

1. Capture data which follows a specific format

   The regular expression itself is then trivially (\d{2})/(\d{2})/(\d{4}) - the point is to check the format, not the validity, and capture the relevant data.

2. Validate the captured data

   int year = Integer.parse(m.groups(3));
   // ..
   if (!validDate(year, month, day)) { .. }
   

Imagine how terrible it would be to try and also validate leap years with a regular expression alone; yet doing so with the above capture-validate is relatively trivial.

¹ This is a (Y) answer, because it assumes that actual code is also allowed. However, the "real life" correct solution (Z) would be to use existing date parsing support in Java or Joda Time, etc.

tl;dr

Regex is overkill for that purpose.

Parse your input using LocalDate class with a DateTimeFormatter, trapping for DateTimeParseException to detect faulty input.

java.time

Java 8+ comes with industry-leading date-time classes found in the java.time packages.

For a date-only (year-month-day) value, use java.time.LocalDate class.

Specify a formatting pattern to match your expected input.

DateTimeFormatter f = DateTimeFormatter.ofPattern( "MM/dd/uuuu" ) ;

Parse your input string as a LocalDate object. To detect faulty input, trap for DateTimeParseException.

try
{
    LocalDate ld = LocalDate.parse( input , f ) ;
    … 
}
catch ( DateTimeParseException e ) 
{
    … handle faulty input …
}