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: [EMAIL PROTECTED] Web Site: http://www.sonsothunder.com/ > -----Original Message----- > From: [EMAIL PROTECTED] > [mailto:[EMAIL PROTECTED] 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 > [EMAIL PROTECTED] > > _______________________________________________ > use-revolution mailing list > [EMAIL PROTECTED] > http://lists.runrev.com/mailman/listinfo/use-revolution > _______________________________________________ use-revolution mailing list [EMAIL PROTECTED] http://lists.runrev.com/mailman/listinfo/use-revolution
