oDate1 = CDate("22/05/2008")
oDate2 = CDate("17/06/2008")
msgbox oDate1 - oDate2
msgbox oDate1 < oDate2
On Fri, Dec 5, 2008 at 9:36 AM, venkata koduri <[EMAIL PROTECTED]>wrote:
> Hi Rick,
>
> I appreciate you. In this code year>0 then it pass. If I enter 2008 or 08
> both or equal. Right?
> My doubt is if year SHOULD BE 4 CHARACTERS then how does it work??
>
> clarify me.
> Thats why I prefer Regular expression.
> On Fri, Dec 5, 2008 at 7:54 AM, Rick Hall <[EMAIL PROTECTED]>wrote:
>
>>
>> Close, and much better than the regular expressions that allow
>> February 30 and other assorted bad dates.
>>
>> arr = split(date, ""\")
>> /* month out of range [1..12] */
>> If arr(0) > 12 OR arr(0) < 1 Then
>> report error
>> End If
>> /* day too small */
>> If arr(1) < 1 Then
>> report error
>> End If
>> /* there is no year 0, negative numbers are BCE */
>> If arr(2) = 0 Then
>> report error
>> End If
>> /* 30 days hath September ...*/
>> If arr(0) = 1 OR arr(0) = 3 OR arr(0) = 5 OR arr(0) = 7 OR arr(0) = 8
>> OR arr(0) = 10 OR arr(0) = 12 THEN
>> If arr(1) > 31 Then
>> report error
>> End If
>> End If
>> If arr(0) = 4 OR arr(0) = 6 OR arr(0) = 9 OR arr(0) = 11 THEN
>> If arr(1) > 30 Then
>> report error
>> End If
>> End If
>> If arr(0) = 2 THEN
>> If arr(1) > 29 Then
>> report error
>> End If
>> If arr(1) = 29 Then
>> IF /* not Leap Year */ THEN
>> /* left as an exercise; years, arr(2), divisible by 4 are leap years,
>> unless they are divisible by 100 when they are not leap years, unless
>> they are divisible by 400 when they are leap years */
>> report error
>> End If
>> End If
>> End If
>>
>> Detailed and messy. But this is the heart of programming. Your
>> database will choke on bad dates.
>>
>> Rick
>>
>>
>>
>
>
> --
> Regards..
> Venkat
>
>
> >
>
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