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Access 2003 chaeck field to see if it contains a certain pattern

Posted on 2011-03-02
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Last Modified: 2012-05-11
Hi There,

I have a 12 digit field.
I want to be able to check if anywhere in that string it contains 4 consecutive non numeric characters.

eg

1BLT010001   No
2POLK1000    Yes
POL1000        NO
POLK9999      yes

Any Ideas
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Question by:EWHTLC
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5 Comments
 
LVL 120

Assisted Solution

by:Rey Obrero (Capricorn1)
Rey Obrero (Capricorn1) earned 1000 total points
ID: 35017059
you will need a function to do that, place this codes in a regular module
is there a possibility to have more than 4 consecutive Alpha characters


function count4Alpha(str) as boolean
count4Alpha=false
if len(str) < 4 then count4Alpha=False: exit function

dim j, strA as string, xCnt as integer
strA="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
for j=1 to len(str)
      if  instr(strA,mid(str,j,1))>0 then
         xCnt=xcnt +1
         if xcnt>=4 then
             count4Alpha=true
             exit function
         end if
         else
         xCnt=0
     end if
next

end function



you can use the function in a query like this

select [field], count4Alpha([field])
fro tableName
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LVL 11

Accepted Solution

by:
LambertHeenan earned 1000 total points
ID: 35018166
Here is an alternative function that checks for any desired number of alpha characters in a row...


function strNStringAlpha(strInput, nConsecutive) as string
' Checks it there is any string of nConsecutive Alpha characters
' (A-Z, a-z) and returns the string if found
dim nLen as Long
dim n as Long
dim c as String 
dim strResult as string 
dim nAlpha
	nLen = Len(strInput)
	For n = 1 to nLen
		c = Ucase(mid(strInput,n,1))
		If c >="A" and c <= "Z" then
			nAlpha = nAlpha + 1
			strResult = strResult & mid(strInput,n,1)
			if nAlpha = nConsecutive then
				Exit For
			End If
		Else
			nAlpha = 0
			strResult = ""
		End If
	Next N
	strNStringAlpha = strResult
End Function

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Author Comment

by:EWHTLC
ID: 35019666
Both Solutions worked a treat.
2nd one slightly better in that it looked for 4 consecutives.

Brilliant response.

Thanks you for your time

Will split the points.
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Author Closing Comment

by:EWHTLC
ID: 35019676
thanks
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LVL 11

Expert Comment

by:LambertHeenan
ID: 35019912
My pleasure.
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