eshurak
asked on
Count numbers in string using VB 6.0
Hello,
I need to count the number character in a string using vb 6.0/vba. For example
NumberCount("%1.2%) = 2
NumbeCount("%8dsf%") = 1
Any ideas?
I need to count the number character in a string using vb 6.0/vba. For example
NumberCount("%1.2%) = 2
NumbeCount("%8dsf%") = 1
Any ideas?
Add the function below to your project, and use it like this:InputString = "1qw23rt45y6"Debug.Print "There are " & Len(RegExpReplace(InputStr ing, "\D")) & " digits in the string"For more info on the function below, please see my article https://www.experts-exchange.com/Programming/Languages/Visual_Basic/A_1336-Using-Regular-Expressions-in-Visual-Basic-for-Applications-and-Visual-Basic-6.html
Function RegExpReplace(LookIn As String, PatternStr As String, Optional ReplaceWith As String = "", _
Optional ReplaceAll As Boolean = True, Optional MatchCase As Boolean = True, _
Optional MultiLine As Boolean = False)
' Function written by Patrick G. Matthews. You may use and distribute this code freely,
' as long as you properly credit and attribute authorship and the URL of where you
' found the code
' For more info, please see:
' http://www.experts-exchange.com/articles/Programming/Languages/Visual_Basic/Using-Regular-Expressions-in-Visual-Basic-for-Applications-and-Visual-Basic-6.html
' This function relies on the VBScript version of Regular Expressions, and thus some of
' the functionality available in Perl and/or .Net may not be available. The full extent
' of what functionality will be available on any given computer is based on which version
' of the VBScript runtime is installed on that computer
' This function uses Regular Expressions to parse a string, and replace parts of the string
' matching the specified pattern with another string. The optional argument ReplaceAll
' controls whether all instances of the matched string are replaced (True) or just the first
' instance (False)
' If you need to replace the Nth match, or a range of matches, then use RegExpReplaceRange
' instead
' By default, RegExp is case-sensitive in pattern-matching. To keep this, omit MatchCase or
' set it to True
' If you use this function from Excel, you may substitute range references for all the arguments
' Normally as an object variable I would set the RegX variable to Nothing; however, in cases
' where a large number of calls to this function are made, making RegX a static variable that
' preserves its state in between calls significantly improves performance
Static RegX As Object
If RegX Is Nothing Then Set RegX = CreateObject("VBScript.RegExp")
With RegX
.Pattern = PatternStr
.Global = ReplaceAll
.IgnoreCase = Not MatchCase
.MultiLine = MultiLine
End With
RegExpReplace = RegX.Replace(LookIn, ReplaceWith)
End Function
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SOLUTION
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ASKER
Thanks Guys,
I'm gonna go with my code below, but it's inspired by some of the solutions above.
I'm gonna go with my code below, but it's inspired by some of the solutions above.
Function NumberCount(theString As String) As Long
Dim i As Long
For i = 1 To Len(theString)
If IsNumeric(Mid(theString, i, 1)) Then NumberCount = NumberCount + 1
Next i
End Function
Just out of curiosity what happens with numbers with more than one digit?
Or is it actually digits you want to count?
Or is it actually digits you want to count?
ASKER
Imnorie, it's the actual number of digits.
function NumberCount( s as string ) as long
dim q as string, c as long
q=s
c=0
while c<=9
q=replace(q, cstr(c),"")
c=c+1
wend
NumberCount=len(s)-len(q)
end function