Spot on Mark.
if yy>70 then yyyy=1900+yy
else yyyy=2000+yy
d1 = .DateTime~new(yyyy,1,1)
d2 = d1~addDays(ddd-1)
say d2~standardDate
OK, this returns the date in the format yyyymmdd, not yymmdd as per my
initial requirement, but even I may be able sort that out!
Thanks.
Staffan
On Fri, Nov 29, 2013 at 9:00 PM, Staffan Tylen <[email protected]>wrote:
> Hi Mark, thanks, I never thought about that. I'll look into it right away.
>
> Btw, today is Nov 29, 2013 ;)
>
>
> Staffan
>
>
>
> On Fri, Nov 29, 2013 at 8:53 PM, Mark Miesfeld <[email protected]> wrote:
>
>> On Fri, Nov 29, 2013 at 11:17 AM, Staffan Tylen
>> <[email protected]>wrote:
>>
>>
>>> I'm playing around with the date() function and am trying to convert a
>>> date in the format yyddd to yymmdd but I can't find a way to handle the yy
>>> portion of the date. The "D" option supports the ddd portion but the result
>>> shows the current year. How can I make yy be part of the calculation?
>>>
>>
>> Hi Staffan,
>>
>> I'm not much help here, dates are too complex for me. If I can figure
>> out today's date, I'm about at my limit.
>>
>> My first thought though is that the DateTime class may be of more help
>> than the date() function.
>>
>> --
>> Mark Miesfeld
>>
>>
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