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I have input strings in nearly standard ISO 8601, but the minute-of-hour part is omitted (don't ask why, funky data feed out of my control).

2018-01-23T12

How do I parse that while letting the minute-of-hour default to zero?

2018-01-23T12:00

DateTimeFormatter.ISO_LOCAL_DATE_TIME

The default parser for LocalDateTime is DateTimeFormatter.ISO_LOCAL_DATE_TIME. This formatter tolerates optional second-of-minute. The fractional second and the second-of-minute are both set to zero.

LocalDateTime ldt = LocalDateTime.parse( "2018-01-23T12:00" ) ;  // Works.

But omitting the minute-of-hour fails, with exception thrown.

LocalDateTime ldt = LocalDateTime.parse( "2018-01-23T12" ) ;  // Fails.

DateTimeFormatterBuilder

I am aware of the DateTimeFormatterBuilder class and its ability to tolerate optional parts while setting default values.

But I have been able to use it properly. I tried the pattern "uuuu-MM-dd HH" while setting both minute-of-hour and second-of-minute to default to zero.

String input = "2018-01-23T12";

DateTimeFormatterBuilder b = new DateTimeFormatterBuilder().parseDefaulting( ChronoField.MINUTE_OF_HOUR , 0 ).parseDefaulting( ChronoField.SECOND_OF_MINUTE , 0 ).appendPattern( "uuuu-MM-dd HH" );
DateTimeFormatter f = b.toFormatter( Locale.US );
LocalDateTime ldt = LocalDateTime.parse( input , f );
System.out.println( ldt );

Exception thrown:

Exception in thread "main" java.time.format.DateTimeParseException: Text '2018-01-23T12' could not be parsed at index 10

at java.base/java.time.format.DateTimeFormatter.parseResolved0(DateTimeFormatter.java:1988)

at java.base/java.time.format.DateTimeFormatter.parse(DateTimeFormatter.java:1890)

at java.base/java.time.LocalDateTime.parse(LocalDateTime.java:492)

at com.basilbourque.example.App.doIt(App.java:31)

at com.basilbourque.example.App.main(App.java:22)

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  • It's a bit hacky, but have you thought of +":00" to the end of the strings? Commented Mar 7, 2018 at 23:29
  • @Autumnal Yes, but… I'm really trying to learn more about how DateTimeFormatterBuilder works. In other words, I'm more interested in an education than a solution. This particular problem seemed like a good example for a Stack Overflow Question, as none of the other six Questions on DateTimeFormatterBuilder with "optional" or "default" cover this. Commented Mar 7, 2018 at 23:41

1 Answer 1

Reset to default
1

If you look carefully at the error message, you will see that it says: Text '2018-01-23T12' could not be parsed at index 10 This means that the problem is the T in the format.

If we then go back to the DateTimeFormatterBuilder, the pattern specified is: "uuuu-MM-dd HH". Here is the problem, this specifies a space when there is in fact a T. The solution is to replace this pattern with: "uuuu-MM-dd'T'HH"

String input = "2018-01-23T12";

DateTimeFormatterBuilder b = new DateTimeFormatterBuilder().parseDefaulting( ChronoField.MINUTE_OF_HOUR , 0 ).parseDefaulting( ChronoField.SECOND_OF_MINUTE , 0 ).appendPattern( "uuuu-MM-dd'T'HH" );
DateTimeFormatter f = b.toFormatter( Locale.US );
LocalDateTime ldt = LocalDateTime.parse( input , f );
System.out.println( ldt );

2018-01-23T12:00

Indeed, you do not even need the builder at all. Specify a DateTimeFormatter with that same pattern. The minute-of-hour does default to zero.

String input = "2018-01-23T12";

DateTimeFormatter f = DateTimeFormatter.ofPattern( "uuuu-MM-dd'T'HH" ) ;
LocalDateTime ldt = LocalDateTime.parse( input , f );

2018-01-23T12:00

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1 Comment

Oh, whoops. In my experimenting, I messed that up. Thanks.

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