Kinda a toughy, take a shot.

[System]

If someone inputs a word in text1, I need text2 to output a new word for every letter entered in text 1.

So If someone enters:
"Hi" in text1,

I need text2 to output:
"ManHam"

So that "H"=Man, and "i"=Ham, and so on...
But I also need "H" to be a different word than "h".

All I really need is the code and I will fill in the word that each letter means.

Thanks if you can help.
-System

[James Stanich]

Here is how I'd do it, but may not be easiest way...

Load values into 2 column array on form load (column 1 being a-z column 2 being the values.)

If only letters, disable keypress of all other cahracters in textbox1. On keypress in textbox1,do a for loop through array to find the letter entered. return value found in textbox2.

Does this make sense?

[gwdash]

Private Sub Text1_KeyPress(KeyAscii As Integer)
Text2.SelStart = Len(Text2)
Select Case UCase(Chr(KeyAscii))
Case "H"
Text2.SelText = "Man"
Case "I"
Text2.SelText = "Ham"
'etc, etc
End Select
End Sub

try This

[System]

gwdash,
That works great, but it is not case sensitive.
I need "H" and "h" to mean two different words.

Thanks,
-System

[gwdash]

Delete the UCase funtion, that makes it Case Sensitive

Private Sub Text1_KeyPress(KeyAscii As Integer)
Text2.SelStart = Len(Text2)
Select Case Chr(KeyAscii)
Case "H"
Text2.SelText = "Man"
Case "I"
Text2.SelText = "Ham"
'etc, etc
End Select
End Sub

[System]

Hey, thanks it works great!

If someone did a backspace in text1, and I wanted text2 to erase the last 5 letters that it ouput, how would I code that?

ie:
If someone did a typo and had to backspace in text1, I would want text2 to erase the last 5 letters in text2, for every 1 letter backspaced in text1.

Thanks,
-System

[HeSaidJoe]

'in the keypress of text1

If keyascii = 8 then text2.text = left(text2.text,5)

[System]

That code sorta works...

I input "Hi", and it output "ManHam", then I backspaced once, and it removed "Ham", but I backspaced again, it would not remove "Man".

So I started testing more, I input "HiHiHi" and it output "ManHamManHamManHam", then I backspaced once and it removed "HamManHamManHam", so I was left with "Man", when it was only suppost to remove once "Ham" off the end.

Know whats wrong?
-System

[PaulLewis]

You should consider going with James's suggestion. You do not need to handle any backspaces or anything by building the Text2 string each time Text1 changes.

The time it takes to figure out what text2 should say after every keystroke in text1 is miniscule and it will save you alot of hassle in working out what keystrokes to handle etc.

This little piece of code should get you going.

Option Explicit
Const numWords = 2
Dim myWords(numWords) As String
Dim myChars(numWords) As String

Private Sub Form_Load()
  
  myChars(1) = "H"
  myWords(1) = "Man"
  
  myChars(2) = "i"
  myWords(2) = "Ham"
  
  
End Sub

Private Sub Text1_Change()
  Text2.Text = GetText(Text1.Text)
End Sub

Private Function GetText(myText As String) As String
  Dim c, index As Integer
  Dim myChar, newText As String
  On Error Resume Next
  For c = 1 To Len(myText)
    myChar = Mid(myText, c, 1)
    index = getIndex(myChar)
    If index = -1 Then
      ' this means that there was no matching word found
      ' you could ignore the letter, or maybe do something else like me
      newText = newText + "Err"
    Else
        newText = newText + myWords(index)
    End If
  Next
  GetText = newText
End Function

Private Function getIndex(ByVal myChar As String) As Integer
  Dim c As Integer
  For c = 1 To numWords
    If myChars(c) = myChar Then
      getIndex = c
      Exit Function
    End If
  Next
  ' index of -1 is our flag to say not found
  getIndex = -1
  
End Function

Regards
Paul Lewis

[System]

Thanks, I put in the code that you supplied, and changed all the text1's, text2's, and form name to the right ones, etc... but I keep getting a hightlighted yellow error at *** which I taged below:

Private Sub frmMain_Load()

myChars(1) = "H"
myWords(1) = "Man"

myChars(2) = "i"
myWords(2) = "Ham"

End Sub

Private Sub text1_Change()
text2.text = GetText(text1.text)
End Sub

***
Private Function GetText(myText As String) As String
***
Dim c, index As Integer
Dim myChar, newText As String
On Error Resume Next
For c = 1 To Len(myText)
myChar = Mid(myText, c, 1)
index = getIndex(myChar)
If index = -1 Then
' this means that there was no matching word found
' you could ignore the letter, or maybe do something else like me
newText = newText + "Err"
Else
newText = newText + myWords(index)
End If
Next
GetText = newText
End Function

Private Function getIndex(ByVal myChar As String) As Integer
Dim c As Integer
For c = 1 To numWords
If myChars(c) = myChar Then
getIndex = c
Exit Function
End If
Next
' index of -1 is our flag to say not found
getIndex = -1

End Function

Know whats wrong?

Thanks,
-System

[PaulLewis]

Nope sorry I don't know whats wrong.

Normally VB will only highlight the line containing the error when you click to run the project. VB normally then gives you an error message which you use to understand the cause.

So if VB gave you an error message, how about telling me that as well? I think you will get far better help if you try to solve these things for yourself a little bit first.

By the way, what version of VB do you use please?

Regards
Paul Lewis

[QWERTY]

Try this:

'//NOTE: IT WILL NOT WORK IF THE USER WILL USE THE DELETE KEY. LET ME KNOW IF YOU WANT THAT FUTURE
Option Explicit
Dim LettToWord() As String

Private Sub Form_Load()
    Dim inCounter As Integer
    Dim I As Integer
    
    '//This is a very boring part of the code in which
    '//I tell the program what it should replace a given letter with
    
    For I = 65 To 90 '//A to Z
        ReDim Preserve LettToWord(I)
        Select Case I
        Case 65
            LettToWord(I) = "STRINGFORA"
        Case 66
            LettToWord(I) = "STRINGFORB"
        Case 67
            LettToWord(I) = "STRINGFORC"
        Case 68
            LettToWord(I) = "STRINGFORD"
        Case 69
            LettToWord(I) = "STRINGFORE"
        End Select
        '//and just go on and on until "Z"
    Next I
    
    For I = 97 To 122 '//a to z
        ReDim Preserve LettToWord(I)
        Select Case I
        Case 97
            LettToWord(I) = "STRINGFORa"
        Case 98
            LettToWord(I) = "STRINGFORb"
        Case 99
            LettToWord(I) = "STRINGFORc"
        Case 100
            LettToWord(I) = "STRINGFORd"
        Case 101
            LettToWord(I) = "STRINGFORe"
        '//and you would have to just go on and on until you've reached "z"
        End Select
    Next I
    
End Sub

Private Sub Text1_KeyPress(KeyAscii As Integer)
    Dim LastChar As String
    
    LastChar = Right(Text1.Text, 1)
    
    If KeyAscii = vbKeyBack Then
        Text2.Text = Left(Text2.Text, Len(Text2.Text) - Len(LettToWord(Asc(LastChar))))
    Else
        Text2.Text = Text2.Text & LettToWord(KeyAscii)
    End If
End Sub

Let me know if something is wrong

[Edited by QWERTY on 09-13-2000 at 09:38 PM]

[System]

Well, If you couldn't tell, I have the "Learning Edition"
I am very new at this, and all the stuff that came with the package is not helping my much. I really do apprechieate(spelling?) you guys's help.

QWERTY:
I am getting some error, i changed all the Text1s' and Text2s' to the correct names, but I am getting this error when I do a test run:

Compile Error:
Sub or Function not defined

It highlights this in yellow:

Private Sub Text1_KeyPress(KeyAscii As Integer)
*note: Text1 was replaced with 'inputtext'

-System

[QWERTY]

Do you have this line in the form's declaration section:

Dim LettToWord() As String

It won't work without it (to make this code work you have to use ALL of it, it won't work if you will just copy a part of it)

[System]

Well, as my General Declarations, I have:
Dim tString As String, tText
Dim LettToWord() As String
Dim LastChar As String
Dim inCounter As Integer
Dim I As Integer

when i run it, i get a
"Run time error '9'
subscript out of range"

and when u click debug, it highlights theline below in yellow:
two.text = two.text & LettToWord(KeyAscii)

So you can see what I have so far, I posted it below
(text1=one, text2=two, and my gen. decs. are above, ):

Option Explicit
Const numWords = 2
Dim myWords(numWords) As String
Dim myChars(numWords) As String
Private Sub frmMain_Load()

'//This is a very boring part of the code in which
'//I tell the program what it should replace a given letter with

For I = 65 To 90 '//A to Z
ReDim Preserve LettToWord(I)
Select Case I
Case 65
LettToWord(I) = "STRINGFORA"
Case 66
LettToWord(I) = "STRINGFORB"
Case 67
LettToWord(I) = "STRINGFORC"
Case 68
LettToWord(I) = "STRINGFORD"
Case 69
LettToWord(I) = "STRINGFORE"
End Select
'//and just go on and on until "Z"
Next I

For I = 97 To 122 '//a to z
ReDim Preserve LettToWord(I)
Select Case I
Case 97
LettToWord(I) = "STRINGFORa"
Case 98
LettToWord(I) = "STRINGFORb"
Case 99
LettToWord(I) = "STRINGFORc"
Case 100
LettToWord(I) = "STRINGFORd"
Case 101
LettToWord(I) = "STRINGFORe"
'//and you would have to just go on and on until you've reached "z"
End Select
Next I

End Sub

Private Sub one_KeyPress(KeyAscii As Integer)
LastChar = Right(one.text, 1)
If KeyAscii = vbKeyBack Then
two.text = Left(two.text, Len(two.text) - Len(LettToWord(Asc(LastChar))))
Else
two.text = two.text & LettToWord(KeyAscii)
End If
End Sub

also, How would I add in the delete key?

Thanks,
-System

[PaulLewis]

It's not that I am the sort that always thinks he's right..far from it. However I have to make the observation here that the approach System is following is prone to complexities not necessary.

There is no need to concern ourselves with the KeyPress event (except as lesson perhaps) if in fact the higer level Change event gives us what we want.

I would normally not post a message like this but in this case I truly beliese System is being led down the wrong path. I hope you are able to see what I am talking about - or perhaps I have missed something System needs to do that requires KeyPress to be used.

Regards

[HeSaidJoe]

system:
you could not have copied Paul's code correctly because I jjust copied it and it works just fine.

I then copy your code and it goes off to never never land.

Perhaps you should give it another shot. It works very well.

Job well done Paul

my little quirk should have been:

If KeyAscii = 8 Then Text2 = Left(Text2, (Len(Text2) - 3))

[Edited by HeSaidJoe on 09-17-2000 at 06:03 PM]

[QWERTY]

System:
What letters are you typing into the textbox1??? It will work only for letters a-e and A-E. You would have to extend the LettToWord array to make it work for all letters
If you have only this part of the code:

    For I = 65 To 90 '//A to Z
        ReDim Preserve LettToWord(I)
        Select Case I
        Case 65
            LettToWord(I) = "STRINGFORA"
        Case 66
            LettToWord(I) = "STRINGFORB"
        Case 67
            LettToWord(I) = "STRINGFORC"
        Case 68
            LettToWord(I) = "STRINGFORD"
        Case 69
            LettToWord(I) = "STRINGFORE"
        End Select
        '//and just go on and on until "Z"
    Next I
    
    For I = 97 To 122 '//a to z
        ReDim Preserve LettToWord(I)
        Select Case I
        Case 97
            LettToWord(I) = "STRINGFORa"
        Case 98
            LettToWord(I) = "STRINGFORb"
        Case 99
            LettToWord(I) = "STRINGFORc"
        Case 100
            LettToWord(I) = "STRINGFORd"
        Case 101
            LettToWord(I) = "STRINGFORe"
        '//and you would have to just go on and on until you've reached "z"
        End Select
    Next I

and you try to type "f" then you will get the error, because I didn't tell the program what I want to replace the f with. You have to do it in the above Sub (I'm quite sure that this is the problem, because that code works just perfect on my computer)