java.time.format.DateTimeParseException: Text could not be parsed at index "dd.MM.yyyy HH:mm"

Question

My code is:

import java.time.LocalDateTime;
import java.time.format.DateTimeFormatter;

DateTimeFormatter formatter = DateTimeFormatter.ofPattern("dd.MM.yyyy HH:mm");
String filterValue = "01.03.2021 0:01";
LocalDateTime parsedDate = LocalDateTime.parse(filterValue, formatter);

I get the exception java.time.format.DateTimeParseException: Text '01.03.2021 0:01' could not be parsed at index 11

or this pattern: `"dd.MM.yyyy H:mm"`. By the way, index 11 is the `0`, use only one if you are parsing values with single-digit hours of day. — deHaar, Aug 20 '21 at 10:56
Indexes in Java are 0-based, so index 11 is `0`, not the space character. — Mark Rotteveel, Aug 20 '21 at 10:58

score 1 · Answer 1 · answered Aug 20 '21 at 11:02

1

The format says "HH" but the value provided is "0". "01.03.2021 00:01" works.

I think its pretty unpractical to force a leading zero for the hour but well, i got a solution.

answered Aug 20 '21 at 11:02

Jens Kisters

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use the pattern suggested by deHaar in the question "dd.MM.yyyy H:mm" – derkoe Aug 20 '21 at 11:03
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with `HH` you are explicitly requesting for the leading `0`. use just one `H` if you don't find practical to have a leading `0`. – MarcoLucidi Aug 20 '21 at 11:05

java.time.format.DateTimeParseException: Text could not be parsed at index "dd.MM.yyyy HH:mm"

1 Answers1