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Add a dayOfYear option #414
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Which one is it? You can construct the value (though the API isn't nice) today: LocalDate(year, 1, 1).plus(dayOfYear - 1, DateTimeUnit.DAY) Parsing from strings, on the other hand, is only available via dirty hacks. |
It is actually for parsing. I have a date time in the format of "yyDDDHHmm". I have been using the format helpers to do most of my parsing thus far. Perhaps the string based version might handle this? Is this - dayOfYear - something we think could make it in a future version? |
Okay, I did try to use the fallback of
but it then gives an assert with message:
So I think it would be useful and a nice way to round out the parsers. :-) |
I need to parse / construct a Julian date, which requires the day of year value. It would fall in nicely with the dayOfMonth and dayOfWeek values. :-)
Or, is there a way I can construct my own WithDate extension that would allow me to add it on my end?
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