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From: Dave on 7 Mar 2006 11:04 "Damon" <nonsense(a)nononsense.com> wrote in message news:XKgPf.63723$Dn4.32072(a)newsfe3-gui.ntli.net... > Hi, > > I have placed this under trim because if I used "Range = > RTrim(Left(!m_add1, 7))" then that would bring back 129-179 but if it was > 69-79 that would bring back 69-79 R. I wanted to be able to trim the r > off. Also the reason I am doing the sum in the first place is to match it > up with the recordcount to see if they are CONTINUOUS, if the sum is less > than the recordcount I can work out if they are "EVEN" or "ODD". OK, the suggestion I gave is still valid Air Code dim s() as string dim i as integer dim t as string dim c as string for i = 1 to len(!m_add1) c = mid$(!m_add1,i,1) if isnumeric(c) then t =t & c else t = t & " " end if next s = split(trim$(t)," ") you now have the two numbers (as strings) in the array s This will work for any delimiter and if there is only a single number it will just return one element in the array. This is just one rough & ready method, given more information about the source data and its consistancy you should be able to come up with something much better. Dave O.
From: Mike Williams on 7 Mar 2006 11:36 "Damon" <nonsense(a)nononsense.com> wrote in message news:XKgPf.63723$Dn4.32072(a)newsfe3-gui.ntli.net... > I can work out if they are "EVEN" or "ODD". If the format of the address is something like the format shown, and if you assume that there are only even numbers (or only odd numbers) in a given address range then you could do something like: Range = (Val(Mid$(s1, InStr(s1, "-") + 1)) - Val(s1)) / 2 + 1 The above should return how many individual houses are included in the range. If the format is likely to be different in some respects then you will of course need to write a bit more code. Mike
From: Mike Williams on 7 Mar 2006 11:38
"Damon" <nonsense(a)nononsense.com> wrote in message news:XKgPf.63723$Dn4.32072(a)newsfe3-gui.ntli.net... .. . . unless of course you have an address range like: 129 - 179 4th Avenue" ;-) Mike |