Unknown pattern letter: T - Parse string date with pattern T to LocalDateTime

Question

I need to parse the following date format in String to Java LocalDateTime.

So I get date as String like this: 2019-09-20T12:36:39.359

I have the following unit test:

@Test
public void testDateTime() {
    assertEquals(SomeObject.getLocalDate(), LocalDateTime.parse(“2019-09-20T12:36:39.359”, DateTimeFormatter.ofPattern("yyyy-MM-ddThh:mm:ss.SSS")));
}

The unit test fails with exception:

java.lang.IllegalArgumentException: Unknown pattern letter: T

    at java.time.format.DateTimeFormatterBuilder.parsePattern(DateTimeFormatterBuilder.java:1661)
    at java.time.format.DateTimeFormatterBuilder.appendPattern(DateTimeFormatterBuilder.java:1570)
    at java.time.format.DateTimeFormatter.ofPattern(DateTimeFormatter.java:536)

How can I correctly parse the date in this format to LocalDateTime?

you can escape the T with single quotes: "yyyy-MM-dd'T'hh:mm:ss.SSS" — Carlos, Oct 01 '19 at 10:32
Your string is in [ISO 8601](https://en.wikipedia.org/wiki/ISO_8601) format, the default for `LocalDateTime`, so you can just leave out the formatter: `LocalDateTime.parse("2019-09-20T12:36:39.359")`. The parsed `LocalDateTime` object won’t ever be equal to a `String`, though. — Ole V.V., Oct 01 '19 at 13:26

score 21 · Answer 1 · answered Oct 01 '19 at 10:42

21

You can also use DateTimeFormatter.ofPattern as below

    DateTimeFormatter dtf = DateTimeFormatter.ofPattern("yyyy-MM-dd'T'HH:mm:ss.SSS", Locale.getDefault());

    String dateStr = "2019-09-20T12:36:39.359";

    LocalDateTime date = LocalDateTime.parse(dateStr, dtf);

answered Oct 01 '19 at 10:42

Sachini Wickramaratne

529
3
17

No need to specify a formatting pattern. This input is in standard ISO 8601 format used by default when parsing/generating text. `LocalDateTime.parse( "2019-09-20T12:36:39.359" )` – Basil Bourque Oct 01 '19 at 15:51

score 5 · Answer 2 · answered Oct 01 '19 at 10:34

5

You can use DateTimeFormatter.ISO_LOCAL_DATE_TIME as the formatter:

LocalDateTime.parse("2019-09-20T12:36:39.359", DateTimeFormatter.ISO_LOCAL_DATE_TIME);

answered Oct 01 '19 at 10:34

Mushif Ali Nawaz

3,707
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18
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How can we get the nanosecond to be exact and strip out the 0s `java.lang.AssertionError: Expected :2019-09-20T12:36:39.359 Actual :2019-09-20T12:36:39.000000359` – M06H Oct 01 '19 at 11:31
@M06H Here you are comparing two `String` values. You can easily strip the `0`s from the `String`. – Mushif Ali Nawaz Oct 01 '19 at 12:14
Ok and how about if it is localdatetime I compare instead of string? – M06H Oct 01 '19 at 12:44

Paplusc · Answer 3 · 2019-10-01T10:46:48.637

5

You are comparing a String with a Date, that will say you they are not equals.

You don't even need to write the DateTimeFormatter.

Writing this code would be enough:

assertEquals("2019-09-20T12:36:39.359", LocalDateTime.parse("2019-09-20T12:36:39.359").toString());

edited Oct 01 '19 at 10:46

answered Oct 01 '19 at 10:34

Paplusc

1,080
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1

The default `DateTimeFormatter` is `DateTimeFormatter.ISO_LOCAL_DATE_TIME`. – Mushif Ali Nawaz Oct 01 '19 at 12:10

Unknown pattern letter: T - Parse string date with pattern T to LocalDateTime

3 Answers3