Excellent Staffan.
--
Mark Miesfeld
On Fri, Nov 29, 2013 at 12:27 PM, Staffan Tylen <[email protected]>wrote:
> 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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>>
>
>
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