SimpleDateFormat cannot parse milliseconds with more than 4 digits

Question

I want to parse a timestamp, like this - \"2016-03-16 01:14:21.6739\". But when I use the SimpleDateFormat to parse it, I find that it outputs an i

Answer 1

tl;dr

LocalDateTime.parse(
    "2016-03-16 01:14:21.6739".replace( " " , "T" )  // Comply with ISO 8601 standard format.
)

Milliseconds versus Microseconds

As others noted, java.util.Date has millisecond resolution. That means up to 3 digits of a decimal fraction of second.

You have 4 digits in your input string, one too many. Your input value demands finer resolution such as microseconds or nanoseconds.

java.time

Instead of using the flawed, confusing, and troublesome java.util.Date/.Calendar classes, move on to their replacement: the java.time framework built into Java 8 and later.

The java.time classes have a resolution of nanosecond, up to nine digits of decimal fraction of a second. For example:

2016-03-17T05:19:24.123456789Z

ISO 8601

Your string input is almost in standard ISO 8601 format used by default in java.time when parsing/generating textual representations of date-time values. Replace that space in the middle with a T to comply with ISO 8601.

String input = "2016-03-16 01:14:21.6739".replace( " " , "T" );

Unzoned

A LocalDateTime is an approximation of a date-time, without any time zone context. Not a moment on the timeline.

LocalDateTime ldt = LocalDateTime.parse( input );

UTC

Make that LocalDateTime an actual moment on the timeline by applying the intended time zone. If meant for UTC, make an Instant.

Instant instant = ldt.toInstant( ZoneOffset.UTC );

Zoned

If meant for a particular time zone, specify a ZoneId to get a ZoneDateTime.

ZoneId zoneId = ZoneId.of( "America/Montreal" );
ZonedDateTime zdt = ldt.atZone( zoneId );