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

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?

If you want to parse all valid formats of offsets (Z, ±hh:mm, ±hhmm and ±hh), one alternative is to use a java.time.format.DateTimeFormatterBuilder with optional patterns (unfortunatelly, it seems that there's no single pattern letter to match them all):

DateTimeFormatter formatter = new DateTimeFormatterBuilder()
    // date/time
    .append(DateTimeFormatter.ISO_LOCAL_DATE_TIME)
    // offset (hh:mm - "+00:00" when it's zero)
    .optionalStart().appendOffset("+HH:MM", "+00:00").optionalEnd()
    // offset (hhmm - "+0000" when it's zero)
    .optionalStart().appendOffset("+HHMM", "+0000").optionalEnd()
    // offset (hh - "Z" when it's zero)
    .optionalStart().appendOffset("+HH", "Z").optionalEnd()
    // create formatter
    .toFormatter();
System.out.println(OffsetDateTime.parse("2022-03-17T23:00:00.000+0000", formatter));
System.out.println(OffsetDateTime.parse("2022-03-17T23:00:00.000+00", formatter));
System.out.println(OffsetDateTime.parse("2022-03-17T23:00:00.000+00:00", formatter));
System.out.println(OffsetDateTime.parse("2022-03-17T23:00:00.000Z", formatter));

All the four cases above will parse it to 2022-03-17T23:00Z.

You can also define a single string pattern if you want, using [] to delimiter the optional sections:

// formatter with all possible offset patterns
DateTimeFormatter formatter = DateTimeFormatter
    .ofPattern("yyyy-MM-dd'T'HH:mm:ss.SSS[xxx][xx][X]");

This formatter also works for all cases, just like the previous formatter above. Check the javadoc to get more details about each pattern.

Notes:

• A formatter with optional sections like the above is good for parsing, but not for formatting. When formatting, it'll print all the optional sections, which means it'll print the offset many times. So, to format the date, just use another formatter.
• The second formatter accepts exactly 3 digits after the decimal point (because of .SSS). On the other hand, ISO_LOCAL_DATE_TIME is more flexible: the seconds and nanoseconds are optional, and it also accepts from 0 to 9 digits after the decimal point. Choose the one that works best for your input data.

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.