Help with Counting Characters

katie_walsh · #1 (**permalink**) 11-15-10, 12:27

Is there anyway I can get access to count the letters in a memo field and not symbols or spaces?

Thanks

Katie

Missinglinq · #2 (**permalink**) 11-15-10, 12:44

There are several ways to do this. This code, in the Click Event of a button, or in some other suitable event, will give you a count of characters without Spaces, Periods, Commas or Exclamation Points

Code:

Dim AlphaOnly As String
  
 If Nz(Me.FieldName, "") <> "" Then
  AlphaOnly = Me.FieldName
  AlphaOnly = Replace(AlphaOnly, " ", "") 'Dump Spaces
  AlphaOnly = Replace(AlphaOnly, ".", "") 'Dump Periods
  AlphaOnly = Replace(AlphaOnly, ",", "") 'Dump Commas
  AlphaOnly = Replace(AlphaOnly, "!", "") 'Dump Exclamation Points
  MsgBox Len(AlphaOnly)
 Else
  MsgBox "Field is Empty!"
 End If

You can add code in a like manner for other symbols that need to be excluded.

Linq ;0)>

katie_walsh · #3 (**permalink**) 11-15-10, 12:56

Thanks, I thought there'd be a way!

Missinglinq · #4 (**permalink**) 11-15-10, 13:40

Another way would be:

Code:

Dim intCounter as Integer
Dim AlphaOnlyCount As Integer

AlphaOnlyCount = 0

If Nz(Me.FieldName, "") <> "" Then
For intCounter = 1 To Len(Me.FieldName)
If Mid$(Me.FieldName, intCounter, 1) Like "[a-z]" Then
 AlphaOnlyCounter = AlphaOnlyCounter + 1
End If
Next
 MsgBox AlphaOnlyCounter
Else
 MsgBox "Field is Empty!"
End If

katie_walsh · #5 (**permalink**) 11-16-10, 06:14

Thanks, that second one looks slightly easier. I don't suppose there would be anyway to get it not to count letters between brackets? I'm trying to get counts of latin inscription lengths but the abbreviations are expanded i.e. 'h s est' becomes 'h(ic) s(itus) est'. I'm beginning to think the easiest thing to do would be to just count the letters I want by hand!

pootle flump · #6 (**permalink**) 11-16-10, 06:43

It would help if you could work out ALL the rules before requesting a solution. Any new rule might invalidate all the work done so far.

Anyway, I think regex might be your best bet. If you used this function from the web (simply copy and paste) then the following works:

Code:

?len(rgxReplace(rgxReplace("This is the string (here) and more", "[^a-z]"), "\(.+\)"))

Result: 26

pootle flump · #7 (**permalink**) 11-16-10, 06:45

Oops - hang on - my regex expression is wrong.

pootle flump · #8 (**permalink**) 11-16-10, 07:03

Fixed:

Code:

?len(rgxReplace(rgxReplace("This is the string (here) and more (another bracket)", "\([^\(\)]+\)", , , True), "[^a-z]", , , True))

returns 22.

katie_walsh · #9 (**permalink**) 11-16-10, 07:48

Yeah, sorry about that. I thought I had...then realised I'd forgotten to take account of the expansions. Thanks for the help.

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