Java 8 Date and Time: parse ISO 8601 string without colon in offset

We try to parse the following ISO 8601 DateTime String with timezone offset:

final String input = "2022-03-17T23:00:00.000+0000";

OffsetDateTime.parse(input);
LocalDateTime.parse(input, DateTimeFormatter.ISO_OFFSET_DATE_TIME);

Both approaches fail (which makes sense as OffsetDateTime also use the DateTimeFormatter.ISO_OFFSET_DATE_TIME) because of the colon in the timezone offset.

java.time.format.DateTimeParseException: Text '2022-03-17T23:00:00.000+0000' could not be parsed at index 23

But according to Wikipedia there are 4 valid formats for a timezone offset: 

<time>Z 
<time>±hh:mm 
<time>±hhmm 
<time>±hh

Other frameworks/languages can parse this string without any issues, e.g. the Javascript Date() or Jacksons ISO8601Utils (they discuss this issue here)

Now we could write our own DateTimeFormatter with a complex RegEx, but in my opinion the java.time library should be able to parse this valid ISO 8601 string by default as it is a valid one.

For now we use Jacksons ISO8601DateFormat, but we would prefer to use the official date.time library to work with. What would be your approach to tackle this issue?

Jon Skeet
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You don't need to write a complex regex - you can build a DateTimeFormatter that will work with that format easily:

DateTimeFormatter formatter =
    DateTimeFormatter.ofPattern("uuuu-MM-dd'T'HH:mm:ss.SSSX", Locale.ROOT);

OffsetDateTime odt = OffsetDateTime.parse(input, formatter);

That will also accept "Z" instead of "0000". It will not accept "+00:00" (with the colon or similar. That's surprising given the documentation, but if your value always has the UTC offset without the colon, it should be okay.

people

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