问如何使@JsonFormat中的毫秒可选用于与Jackson进行时间戳解析？

EN

JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss[.SSS]X")

JsonFormat(shape = JsonFormat.Shape.STRING, pattern = "yyyy-MM-dd'T'HH:mm:ss[.SSS][.SS][.S]X")

class DateDeserializer extends StdDeserializer<Date> {

    private static final SimpleDateFormat withMillis = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ss.SSSX");
    private static final SimpleDateFormat withoutMillis = new SimpleDateFormat("yyyy-MM-dd'T'HH:mm:ssX");

    public DateDeserializer() {
        this(null);
    }

    public DateDeserializer(Class<?> vc) {
        super(vc);
    }

    @Override
    public Date deserialize(JsonParser p, DeserializationContext ctxt) throws IOException, JsonProcessingException {
        String dateString = p.getText();
        if (dateString.isEmpty()) {
            //handle empty strings however you want,
            //but I am setting the Date objects null
            return null;
        }

        try {
            return withMillis.parse(dateString);
        } catch (ParseException e) {
            try {
                return withoutMillis.parse(dateString);
            } catch (ParseException e1) {
                throw new RuntimeException("Unable to parse date", e1);
            }
        }
    }
}

@Provider
class ObjectMapperContextResolver implements ContextResolver<ObjectMapper> {

    private final ObjectMapper mapper;

    public ObjectMapperContextResolver() {
        mapper = new ObjectMapper();
        SimpleModule dateModule = new SimpleModule();
        dateModule.addDeserializer(Date.class, new DateDeserializer());
        mapper.registerModule(dateModule);
    }

    @Override
    public ObjectMapper getContext(Class<?> type) {
        return mapper;
    }
}