Can I convert this code easily?

I've got variables in my program filled with file names from HTML file, that end up looking like this:

"Ed%20Rush%20%26%20Optical"

I want this to look like:

"Ed Rush & Optical"

Is there a function that will convert this for me? Or is there a 'find and replace' function that I could use?

Thanks for the help!!

Matt

ps. If you make a good enough OS (that's still 100% compatible) I know everyone will happily drop windows!!

[Buzby]

Pop this into a module;

Option Explicit

Function ConvString(xStr As String) As String
Dim Dummy As Integer
ConvString = ""
For Dummy = 1 To Len(xStr)
If Mid(xStr, Dummy, 1) = "%" Then
ConvString = ConvString + Chr(HexToDec(Mid(xStr, Dummy + 1, 2)))
Dummy = Dummy + 2
Else
ConvString = ConvString + Mid(xStr, Dummy, 1)
End If
Next
End Function

Function HexToDec(xHex As String) As Integer
HexToDec = 0
HexToDec = CInt(Left(xHex, 1)) * 16
HexToDec = HexToDec + CInt(Right(xHex, 1))
End Function


then you can use the ConvString function to convert;

eg;

MyString = ConvString("Ed%20Rush%20%26%20Optical")

would return

Ed Rush & Optical


It could probably do with a bit of tweaking but it should work (providing all ascii codes that appear after the % signs are two chars long - if they aren't always you will need to fiddle around with the code a bit)

Have fun.

[Stevie]

I've got a find/replace function I have always used.

Public Function String_ReplaceAll(ByVal strSource As String, _
                                  ByVal strWhat As String, _
                                  ByVal strWith As String) As String
'***************************************************************************
'Purpose:       Replaces all occurances of one string with another string.
'Parameters:    strSource - String to be changed.
'               strWhat   - What needs replacing.
'               strWith   - What it should be replaced with.
'Returns:       String    - The changed string.
'***************************************************************************
On Error GoTo ErrorHandler

Dim intstrWhatLength As Integer
Dim intReplaceLength As Integer
Dim intStart As Integer

  intstrWhatLength = Len(strWhat)
  
  If intstrWhatLength = 0 Then
    
    String_ReplaceAll = strSource
    Exit Function
  
  End If
  
  intReplaceLength = Len(strWith)
  intStart = InStr(1, strSource, strWhat)
  
  Do While intStart > 0
      
    strSource = Left(strSource, intStart - 1) + strWith + Right(strSource, Len(strSource) - (intStart + intstrWhatLength - 1))
    intStart = InStr(intStart + intReplaceLength, strSource, strWhat)
  
  Loop
  
  String_ReplaceAll = strSource

End Function

Hope it helps.

[Buzby]

You can't use a standard find and replace because you need to scan for the % sign and convert whatever's after it from hex to decimal, then to a chr of that to get the special character - eg %20 = space, %26 = & sign.

If it was a simple as that you could use the VB6 Replace function.

This should work.

Option Explicit

Private Sub Form_Load()
    MsgBox DeHTML("Ed%20Rush%20%26%20Optical")
End Sub

Private Function DeHTML(ByVal sInput As String)
    Dim lCounter As Long
    For lCounter = 1 To Len(sInput)
        If Mid$(sInput, lCounter, 1) = "%" Then
            DeHTML = DeHTML & Chr("&H" & Mid$(sInput, lCounter + 1, 2))
            lCounter = lCounter + 2
        Else
            DeHTML = DeHTML & Mid$(sInput, lCounter, 1)
        End If
    Next
End Function

PS. Forgot to press submit, and when i did other ppl had posted already

[Edited by Azzmodan on 04-07-2000 at 05:29 PM]

[Buzby]

I don't think that'll work either 'cos the Chr("&H" +... but will always return & !?

[Stevie]

Buzby, I was only trying to help, and I misunderstood what was needed, but do try to remember that not everybody has the latest version of VB.

I don't think that'll work either 'cos the Chr("&H" +... but will always return & !?

Erm i tested it prior to posting in on this forum, and it seemed to work good on my pc..
Chr("&H" + = Chr("&H" &[/i]