Reverse line

[timjg]

Can anyone help with the code for this command button? Thanks.

Add a "Reverse Line" button. Reverse Line should take an entry such as

"This is some text"

and reverse it as follows:

"text some is This".

[Tiovital]

Hi,

Try NewText.Text = ReverseString(Text1.Text)


Regards

[d.paulson]

You forgot to supply your ReverseString function. VB does not have this function built in.

[Chris]

Hi, The ReverseStr function is available in VB, but are you using VB 6.0?

By the way, the ReverseStr may not return what you looking for, because it will return the folloing result:

Source string = "This is some text"
Retrun string = "txet emos si sihT"

So, I have wrote a simple ReverseStr function by using the VBScript 5.0 or higher DLL and the code is show below:

Private Function ReverseStr(ByVal sourceStr As String) As String
Dim regex As RegExp
Dim col As MatchCollection
Dim match As match
Dim str() As String
Dim result As String
Dim length As Long
Dim Idx As Integer

'***************************************************************
'   This example code need the VBScript.dll file
'   Before you use this code make sure you have
'   this DLL in your Windows system folder.
'
'   You can download this DLL from the following URL
'http://www.microsoft.com/msdownload/vbscript/scripting51.asp
'
'***************************************************************

Set regex = New RegExp
With regex
    .IgnoreCase = True
    .Global = True
    '\S* search for all non white space charater
    '\s  search for all space character
    .Pattern = "\S*\s"
    Set col = .Execute(sourceStr)
End With
    
Idx = 0
length = 0
For Each match In col
    ReDim Preserve str(Idx)
    str(Idx) = match.Value
    length = length + match.length
    Idx = Idx + 1
Next

'get the last string
If length <> Len(sourceStr) Then
    ReDim Preserve str(Idx)
    str(Idx) = Mid(sourceStr, length) & Chr(32)
End If

For Idx = UBound(str) To 0 Step -1
    result = result & str(Idx)
Next

If result <> "" Then MsgBox result
    
Set regex = Nothing
End Function

If I'm not correct, shouldn't it be StrReverse rather than ReverseStr?

[Chris]

Ya, Magatron you're rite the function in VB is StrReverse, Sorry for my typing mistake.

Similar to the one above, but a bit different:

Private Sub Command1_Click()

    Dim vArray As Variant
    Dim strArray As String
    Dim iArray As Integer
    
    vArray = Split("This is some text", " ")
    For iArray = 0 To UBound(vArray)
        strArray = strArray & Chr(32) & StrReverse(vArray(iArray))
    Next iArray
    MsgBox StrReverse(strArray)
    
End Sub

[timjg]

Will any of these codings work when the string is taken from a user-entered text box? I need to make it work so that if a user enters some text in a text box and then clicks on a reverse line command button it reverses the string. Should any of the above codings be modified? Thanks in advance.

All code above will work, just find where it is used and change it to the textbox (Text1).

Private Sub Command1_Click()

    Dim vArray As Variant
    Dim strArray As String
    Dim iArray As Integer
    
    vArray = Split(Text1.text, " ")
    For iArray = 0 To UBound(vArray)
        strArray = strArray & Chr(32) & StrReverse(vArray(iArray))
    Next iArray
    MsgBox StrReverse(strArray)
    
End Sub

Or:

Private Sub Command1_Click()
    Dim x As Integer
    Dim intLow As Integer
    Dim intHigh As Integer
    Dim strInput As String
    Dim strTest() As String
    Dim strNew As String
    
    strInput = Text1.text
    strTest = Split(strInput)
    intLow = LBound(strTest)
    intHigh = UBound(strTest)
    
    For x = intHigh To intLow Step -1
        strNew = strNew & strTest(x) & " "
    Next x
    
    MsgBox "Original String = " & strInput & vbCrLf & _
           "Reverse String  = " & strNew
End Sub

Or, I guess sourceStr would be Text1.text:

Private Function ReverseStr(ByVal sourceStr As String) As String
Dim regex As RegExp
Dim col As MatchCollection
Dim match As match
Dim str() As String
Dim result As String
Dim length As Long
Dim Idx As Integer

'***************************************************************
'   This example code need the VBScript.dll file
'   Before you use this code make sure you have
'   this DLL in your Windows system folder.
'
'   You can download this DLL from the following URL
'http://www.microsoft.com/msdownload/vbscript/scripting51.asp
'
'***************************************************************

Set regex = New RegExp
With regex
    .IgnoreCase = True
    .Global = True
    '\S* search for all non white space charater
    '\s  search for all space character
    .Pattern = "\S*\s"
    Set col = .Execute(sourceStr)
End With
    
Idx = 0
length = 0
For Each match In col
    ReDim Preserve str(Idx)
    str(Idx) = match.Value
    length = length + match.length
    Idx = Idx + 1
Next

'get the last string
If length <> Len(sourceStr) Then
    ReDim Preserve str(Idx)
    str(Idx) = Mid(sourceStr, length) & Chr(32)
End If

For Idx = UBound(str) To 0 Step -1
    result = result & str(Idx)
Next

If result <> "" Then MsgBox result
    
Set regex = Nothing
End Function

[Tiovital]

Hi,
Opssss.......Sorry !!!!!! I forgot The Function.
But With All GREAT samples that u have now - i dont think u need my poor sample - But here it is.

MyReversText.text = ReverseString(Text1.Text)

Private Function ReverseString(strSource As String) As String
'
Dim pos As Integer 'position
Dim strDummy As String 'dummy string
Dim intC As Integer 'counter
Const Space As String = " " 'space
'
pos = Len(strSource)
strDummy = ""
For intC = Len(strSource) To 1 Step -1
If Mid(strSource, intC, 1) = Space Then
strDummy = strDummy & Mid(strSource, intC, pos - intC + 1)
pos = intC
End If
Next intC
strDummy = strDummy & Left(strSource, pos - intC)
ReverseString = Right(strDummy, Len(strDummy) - 1)
'
End Function


Regards

[kedaman]

Might be handy if you want some speed

Function fastStrReverse(text As String) As String
Dim a() As Byte, b() As Byte, c As Long, n As Long
    a = StrConv(text, vbFromUnicode)
    c = UBound(a)
    ReDim b(c)
    For n = 0 To c
        b(c - n) = a(n)
    Next n
    fastStrReverse = StrConv(b, vbUnicode)
End Function

Fast as you can get if you have vb5, but in vb6 the built in is probably faster