Java, regular expression for time

Question

I have a textfield in Java eclipse interface for which i want to validate it to accept only time format, but I don't want seconds included.

How do I validate it to accept only this format hh:mm but from hours 8:00 morning till 16:00 afternoon.

P.s TextField variable name is txtOra.

Welcome on Stack Overflow. I recommend you to [take the tour](https://stackoverflow.com/tour) and read the [Help Center](https://stackoverflow.com/help) for information on asking a good and well-received question. You are missing a couple points there. — Ben, May 28 '18 at 12:32
Which graphic framework are you using? Is it swing, awt, or maybe android? Are you sure it is `TextField` or maybe it is `JTextField` (these are not the same and solution may depend on it). — Pshemo, May 28 '18 at 12:36
Possibly related: https://docs.oracle.com/javase/tutorial/uiswing/components/formattedtextfield.html, https://docs.oracle.com/javase/8/docs/technotes/guides/swing/1.4/ftf.html — Pshemo, May 28 '18 at 12:39
[Check this](https://stackoverflow.com/questions/25873636/regex-pattern-for-exactly-hhmmss-time-string) Pattern: ```([8-9]|1[0-6]):([0-5][0-9])``` — Andry, May 28 '18 at 12:40
This question has been asked and answered with variations many times. For example [Regex pattern for EXACTLY HH:MM:SS time String](https://stackoverflow.com/questions/25873636/regex-pattern-for-exactly-hhmmss-time-string). — Ole V.V., May 28 '18 at 13:04

Zarathustra · Answer 1 · 2018-05-28T13:53:08.883

3

Just parse it with a DateTimeFormatter

DateTimeFormatter dateTimeFormatter = DateTimeFormatter.ofPattern("H:mm");

LocalTime localTime = LocalTime.parse("16:16", dateTimeFormatter);

if parse does not throw an exception you have a valid time.

Then check for your constraints with LocalDate#isAfter and LocalDate#isBefore

Look it up here to find more patterns https://docs.oracle.com/javase/8/docs/api/java/time/format/DateTimeFormatter.html

edited May 28 '18 at 13:53

answered May 28 '18 at 12:35

Zarathustra

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I would also validate that the time is within the given range after parsing. – clabe45 May 28 '18 at 13:36

Youcef LAIDANI · Answer 2 · 2018-05-28T13:14:26.417

1

If you are using java.time you can use LocalTime where you can use isBefore and isAfter like so :

public static boolean checkDateIsCorrect(String input){
    //Format your date
    LocalTime time = LocalTime.parse(input, DateTimeFormatter.ofPattern("H:mm"));
    //Then check if the time is between 8:00 and 16:00
    return !time.isBefore(LocalTime.parse("08:00")) 
                || !time.isBefore(LocalTime.parse("16:00"));
}

edited May 28 '18 at 13:14

answered May 28 '18 at 12:39

Youcef LAIDANI

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`H` alone can accept 8 as well as 08 (so you don’t need optional parts). Your method rejects times lying on the limits 08:00 and 16:00, which I suspect wasn’t intended from the asker. – Ole V.V. May 28 '18 at 12:58
@OleV.V. I don't understand *Your method rejects times lying on the limits 08:00 and 16:00, which I suspect wasn’t intended from the asker.* – Youcef LAIDANI May 28 '18 at 13:02
What is the expected result of `checkDateIsCorrect("8:00")`? What is the actual result? Same question for `"16:00"`. – Ole V.V. May 28 '18 at 13:06
@OleV.V. I got it, what about now? – Youcef LAIDANI May 28 '18 at 13:07
If you want that way of solving it, I think you need 16:01. – Ole V.V. May 28 '18 at 13:08
Similar to [my answer here](https://stackoverflow.com/a/50564365/5772882) – Ole V.V. May 28 '18 at 13:11
Yes, your way not need to use `7:59` and `16:01` I like it I will go with that one instead thank you @OleV.V. – Youcef LAIDANI May 28 '18 at 13:15

score 0 · Answer 3 · answered May 28 '18 at 12:38

0

Use the following function to validate a time string with HH:MM (24 Hours Format)

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

public boolean validate(final String time){
      String TIME24HOURS_PATTERN = 
                "([01]?[0-9]|2[0-3]):[0-5][0-9]";
      Pattern pattern = Pattern.compile(TIME24HOURS_PATTERN);
      Matcher matcher = pattern.matcher(time);
      return matcher.matches();

}

answered May 28 '18 at 12:38

nandal

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how would you extend this to include seconds? "([01]?[0-9]|2[0-3]):[0-5][0-9]:[0-5][0-9]" ? – ennth Dec 08 '20 at 15:00
@ennth, solution given by you should work for including seconds – nandal Dec 18 '20 at 16:56

Java, regular expression for time

3 Answers3