Find '~XX~' within a string with specific values - regex

Question

I have classic ASP written in VBScript. I have a record pulled from SQL Server and the data is a string. In this string, I need to find text enclosed in ~12345~ and I need to replace with very specific text. Example 1 would be replaced with M, 2 would be replaced with A. I then need to display this on the web page. We don't know how many items will be enclosed with ~.
Example Data:
Group Pref: (To be paid through WIT)
~2.5~ % Quarterly Rebate - Standard Commercial Water Heaters
Display on webpage after:
Group Pref: (To be paid through WIT)
~A.H~ % Quarterly Rebate - Standard Commercial Water Heaters
I tried this following, but there are two many cases and this would be unrealistic to maintain. I does replace the text and display correctly.
dim strSearchThis
strSearchThis =(rsResults("PREF"))
set re = New RegExp
with re
.global = true
.pattern = "~[^>]*~"
strSearchThis = .replace(strSearchThis, "X")
end with
I am also trying this code, I can find the text contained between each ~ ~, but when displayed its the information between the ~ ~ is not changed:
dim strSearchThis
strSearchThis =(rsResults("PREF"))
Set FolioPrefData = New RegExp
FolioPrefData.Pattern = "~[^>]*~"
FolioPrefData.Global = True
FolioPrefData.IgnoreCase = True
'will contain all found instances of ~ ~'
set colmatches = FolioPrefData.Execute(strSearchThis)
Dim itemLength, found
For Each objMatch in colMatches
Select Case found
Case "~"
'ignore - doing nothing'
Case "1"
found = replace(strSearchThis, "M")
End Select
Next
response.write(strSearchThis)

Answer 1

You can do it without using Regular Expressions, just checking the individual characters and writing a function that handles the different cases you have. The following function finds your delimited text and loops through all characters, calling the ReplaceCharacter function defined further down:
Function FixString(p_sSearchString) As String
Dim iStartIndex
Dim iEndIndex
Dim iIndex
Dim sReplaceString
Dim sReturnString
sReturnString = p_sSearchString
' Locate start ~
iStartIndex = InStr(sReturnString, "~")
Do While iStartIndex > 0
' Look for end ~
iEndIndex = InStr(iStartIndex + 1, sReturnString, "~")
If iEndIndex > 0 Then
sReplaceString = ""
' Loop htrough all charatcers
For iIndex = iStartIndex + 1 To iEndIndex - 1
sReplaceString = sReplaceString & ReplaceCharacter(Mid(sReturnString, iIndex, 1))
Next
' Replace string
sReturnString = Left(sReturnString, iStartIndex) & sReplaceString & Mid(sReturnString, iEndIndex)
' Locate next ~
iStartIndex = InStr(iEndIndex + 1, sReturnString, "~")
Else
' End couldn't be found, exit
Exit Do
End If
Loop
FixString = sReturnString
End Function
This is the function where you will enter the different character substitutions you might have:
Function ReplaceCharacter(p_sCharacter) As String
Select Case p_sCharacter
Case "1"
ReplaceCharacter = "M"
Case "2"
ReplaceCharacter = "A"
Case Else
ReplaceCharacter = p_sCharacter
End Select
End Function
You can use this in your existing code:
response.write(FixString(strSearchThis))

Answer 2

You can also use a Split and Join method...
Const SEPARATOR = "~"
Dim deconstructString, myOutputString
Dim arrayPointer
deconstructString = Split(myInputString, SEPARATOR)
For arrayPointer = 0 To UBound(deconstructString)
If IsNumeric(deconstructString(arrayPointer)) Then
'Do whatever you need to with your value...
End If
Next 'arrayPointer
myOutputString = Join(deconstructString, "")
This does rely, obviously, on breaking a string apart and rejoining it, so there is a sleight overhead on string mutability issues.