Re: Those flippin dates

["rick baird" <rick.baird@xxxxxxxxx>]

i've used this for years, before date fields (I see no reason to make
them a date only to move them back into numbers)

d $FlipDate1              1      6s 0
d $FlipDate2              7     12s 0
d  $ymd                   5     10s 0
d  $mdy                   3      8s 0

c* ymd to mdy

c                 eval      $FlipDate1 = Oldymd
c                 eval      $FlipDate2 = Oldymd
c                 eval      Newmdy = $mdy

c* mdy to ymd

c                 eval      $FlipDate1 = Oldmdy
c                 eval      $FlipDate2 = Oldmdy
c                 eval      Newymd = $mdy



On 9/5/07, Pete Helgren <Pete@xxxxxxxxxx> wrote:
> What would be the most expedient way to convert a 6.0 numeric field that
> represents a date in MMDDYY format to a 6.0 numeric field that
> represents a date in YYMMDD order?  The resultant numeric is used in an
> SQL statement comparing it to a numeric (not a date) field.  The old
> technique of  YYMMDD = MMDDYY * 10000.0001 appears not to work because
> the program is cranky at runtime about the result field not being large
> enough.
>
> Since the compare is a numeric I figured that the math would be
> numeric.  Or, is using a date field and converting it to a numeric host
> variable for the SQL the way to go?
>
> Pete Helgren
>
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