How to parse with SimpleDateFormat correctly?

Question

I have this exception with this code:

java.text.SimpleDateFormat@42ee54java.text.ParseException: Unparseable date: "Fri Dec 25 02:00:00 EET 2020"

@PostMapping
public ResponseEntity<?> addEvent(@Valid @RequestBody Event event) {
    try {
        SimpleDateFormat format1 = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy");
        format1.parse(String.valueOf(event.getDate()));
        format1.setLenient(false);
        eventService.save(event);
        return new ResponseEntity<>(event, HttpStatus.CREATED);
    } catch (IllegalArgumentException | ParseException e) {
        return new ResponseEntity<>(HttpStatus.BAD_REQUEST);
    }
}

Due to date format is defined correctly. What can be wrong here? Thanks in advance.

I recommend you don’t use `SimpleDateFormat` and `Date`. Those classes are poorly designed and long outdated, the former in particular notoriously troublesome. Instead use `ZonedDateTime` and `DateTimeFormatter`, both from [java.time, the modern Java date and time API](https://docs.oracle.com/javase/tutorial/datetime/). — Ole V.V., Mar 23 '20 at 14:08
What is the type returned from `event.getDate()`? Asking because it may look like you are converting a `Date` to a `String` only to parse it back into a `Date`? — Ole V.V., Mar 23 '20 at 19:39

score 3 · Accepted Answer · answered Mar 23 '20 at 12:42

3

You have to set Locale, like this:

SimpleDateFormat format1 = new SimpleDateFormat("EEE MMM dd HH:mm:ss z yyyy", Locale.ENGLISH);

because of Fri Dec in your input string and your default locale is probably not english

answered Mar 23 '20 at 12:42

alex

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How to parse with SimpleDateFormat correctly?

1 Answers1