trjhtr

I have written a VBA function that will do parsing / replacement according to tokens / placeholders.

Example:

Input string: Username %u, date is %d.

The function would replace %d with the current date and %u with the current username; the output would look like:
Username Andy, date is 20170820.

Sounds simple to implement, but here's a twist: Regexps and Replace() are not safe. In the example above, if %u after replacement would contain another token (let's say Andy%d), there will be a recursive replacement and the output will mangled: Username Andy20170820, date is 20170820.

I could write this efficiently in C++ or some other "proper" language but I'm relegated to the VBA. I don't want to work on Chars inside a string as that doesn't look very efficient (and I might be using this formula to parse 10000 lines in an Excel sheet).

Here's my solution; it works, looks good and stable and allows for easy extension / customisation, but I'm not sure I have written it in the most efficient way (and in the best style). Your input would be very much appreciated :)

Function AI_ParseMagicSymbols(ByVal TextToParse As String) As String

'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''
'' Replace magic symbols (placeholders) with dynamic data.
''
'' Arguments: a string full of magic.
''
'' Placeholders consist of one symbol prepended with a %:
'' %d - current date
'' %t - current time
'' %u - username (user ID)
'' %n - full user name (usually name and surname)
'' %% - literal % (placeholder escape)
'' Using an unsupported magic symbol will treat the % literally, as if it had been escaped.
'' A single placeholder terminating the string will also be treated literally.
'' Magic symbols are case-sensitive.
''
'' Returns: A string with no magic but with lots of beauty.
''
'' Examples:
'' "Today is %d" becomes "Today is 2018-01-26"
'' "Beautiful time: %%%t%%" yields "Beautiful time: %16:10:51%"
'' "There are %zero% magic symbols %here%.", true to its message, outputs "There are %zero% magic symbols %here%."
'' "%%% looks lovely %%%" would show "%% looks lovely %%" - one % for the escaped "%%" and the second one for the unused "%"!
''
'' Alexander Ivashkin, 26 January 2018
'''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''''

Dim sFinalResult As String
Dim aTokenizedString() As String
Dim sTempString As String
Dim sPlaceholder As String
Dim sCurrentString As String
Dim iIterator As Integer
Dim iTokenizedStringSize As Integer
Dim bThisStringHasPlaceholder As Boolean

' Default placeholder is "%"
Const cPlaceholderSymbol As String = "%"

aTokenizedString = Split(Expression:=TextToParse, Delimiter:=cPlaceholderSymbol)
iTokenizedStringSize = UBound(aTokenizedString())
bThisStringHasPlaceholder = False
sFinalResult = ""

For iIterator = 0 To iTokenizedStringSize
 sCurrentString = aTokenizedString(iIterator)

 If bThisStringHasPlaceholder Then
 If sCurrentString <> "" Then
 sPlaceholder = Left(sCurrentString, 1)
 sTempString = Right(sCurrentString, Len(sCurrentString) - 1)

 ' This is the place where the MAGIC happens
 Select Case sPlaceholder
 Case "d":
 sCurrentString = Date & sTempString
 Case "t":
 sCurrentString = Time & sTempString
 Case "u":
 sCurrentString = Environ$("Username") & sTempString
 Case "n":
 sCurrentString = Environ$("fullname") & sTempString
 Case Else:
 sCurrentString = cPlaceholderSymbol & sCurrentString
 End Select
 Else
 ' We had two placeholders in a row, meaning that somebody tried to escape!
 sCurrentString = cPlaceholderSymbol
 bThisStringHasPlaceholder = False
 End If
End If

sFinalResult = sFinalResult & sCurrentString

If sCurrentString = "" Or (iIterator + 1 <= iTokenizedStringSize And sCurrentString <> cPlaceholderSymbol) Then
 ' Each string in the array has been split at the placeholders. If we do have a next string, then it must contain a magic symbol.

 bThisStringHasPlaceholder = True
 ' Even though it is called "...ThisString...", it concerns the NEXT string.
 ' The logic is correct as we will check this variable on the next iteration, when the next string will become ThisString.
Else
 bThisStringHasPlaceholder = False
End If

Next iIterator

AI_ParseMagicSymbols = sFinalResult

End Function

My immediate idea, without looking at your code, was to use split. I tried to optimize your code by using some tricks:

• To get rid of the problem with the double %, I replace this by a dummy-string (that will not appear in your input, hope you can accept this).


• I leave the first token alone. Either the input starts with a % - in that case it's empty anyhow, or it doesn't, so we don't have to look at it at all.


• At the end, the dummy-string is replaced back with a single %

I ended up with:

Function AI_ParseMagicSymbols(ByVal s As String) As String
 Const PlaceholderSymbol As String = "%"
 Const ReplaceEscapedChar As String = "<ESCAPEPERCENT>"

 s = Replace(s, PlaceholderSymbol & PlaceholderSymbol, ReplaceEscapedChar)
 Dim tokens() As String, i As Long
 tokens = Split(s, PlaceholderSymbol)

 AI_ParseMagicSymbols = tokens(0) ' We don't treat the first token
 For i = 1 To UBound(tokens)
 Dim token As String, placeHolderChar As String, replaceStr As String
 token = tokens(i)
 placeHolderChar = Left(token, 1)

 ' This is the place where the MAGIC happens
 Select Case placeHolderChar
 Case "d":
 replaceStr = Date
 Case "t":
 replaceStr = Time
 Case "u":
 replaceStr = Environ$("Username")
 Case "n":
 replaceStr = Environ$("fullname")
 Case Else:
 replaceStr = "%" & placeHolderChar ' No magic here, keep % and the first char of token
 End Select

 AI_ParseMagicSymbols = AI_ParseMagicSymbols & replaceStr & Mid(token, 2)
 Next i

 AI_ParseMagicSymbols = Replace(AI_ParseMagicSymbols, ReplaceEscapedChar, PlaceholderSymbol)

End Function