RegEx question

[Ken Ray]

Yves,

As Brian pointed out, you just need to copy what you are doing with the
month for the day... that is, take what Dar suggested:

  (0?[1-9]|[12][0-9]|3[01])

and just take out the ?:

  (0[1-9]|[12][0-9]|3[01])

This will enforce the need for leading zeroes.

Ken Ray
Sons of Thunder Software
Email: kray at sonsothunder.com
Web Site: http://www.sonsothunder.com/



> -----Original Message-----
> From: use-revolution-bounces at lists.runrev.com 
> [mailto:use-revolution-bounces at lists.runrev.com] On Behalf Of 
> Yves COPPE
> Sent: Saturday, April 10, 2004 7:12 AM
> To: How to use Revolution
> Subject: Re: RegEx question
> 
> 
> Hi Dar
> 
> 
> 
> > Maybe this will work:
> >
> > function CheckDate pDateToCheck
> >   return
> > 
> matchText(pDateToCheck,"\A(0?[1-9]|[12][0-9]|3[01])/(0[1-9]|1[0-2])/ 
> > ([0-9][0-9][0-9][0-9])\z")
> > end CheckDate
> >
> > It is not consistent yet as to whether the leading zero is 
> optional.   
> > It is optional for the day of the month, but required for the month.
> >
> 
> 
> i'm almost on the good regEx
> but now I'd like that the day MUST be two chars and not an optional
> 
> so :
> 
> 05/04/2004   returns true
> and
> 5/04/2004  should  return false (at this moment, it returns true)
> 
> Can you give me one more hint ??
> 
> Amicalement.
> 
> Yves COPPE
> yvescoppe at skynet.be
> 
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