Splitting a string into variables? - Xtreme Visual Basic Talk

03-01-2003, 08:11 PM

AcidGoat

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Splitting a string into variables?

Ok I found some threads that sort of touched on this, but none that really gave a straight answer. I have a string of characters and I want each of those characters to be a seperate variable. For example, say I have a string that says "Bob". I want one variable for "B", one for "o" and one for "b" So that I can deal with each seperately.

Thanks in advance

--AcidGoat

03-01-2003, 08:17 PM

Kaluriel

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Ostr is the old string
Nstr is the new string

Code:

ReDim Nstr(Len(Ostr))
For i = 1 To Len(Ostr)
    Nstr(i) = Mid(Ostr, i, 1)
Next i

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03-01-2003, 08:21 PM

Squirm

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Or..

Code:

Dim b() As Byte
b = StrConv(OStr, vbFromUnicode)

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03-01-2003, 08:23 PM

Kaluriel

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**** you, your always one step ahead lol

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03-01-2003, 08:29 PM

AcidGoat

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Hmm... maybe I should have mentioned this: the string will always be 12 characters long. I need each character to be its own variable, appropriately named. The first character will be a variable called "char1" the second "char2" etc... Example: say the string is "goatgoatgoat" (excuse the lack of creativity

). I want this to result: char1 = g, char2 = o, char3 = a, char4 = t, char5 = g, char6 = o, char7 = a, char8 = t, char9 = g, char10 = o, char11 = a, char12 = t.

Is this even possible?

03-01-2003, 08:41 PM

Machaira

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Sure, but it'll be ugly.

Code:

Dim char1 As String
    Dim char2 As String
    Dim char3 As String
    Dim char4 As String
    Dim char5 As String
    Dim char6 As String
    Dim char7 As String
    Dim char8 As String
    Dim char9 As String
    Dim char10 As String
    Dim char11 As String
    Dim char12 As String
    
    Dim sString As String
    
    Dim iLp As Integer
    
    sString = "goatgoatgoat"
    
    Dim sChar As String
    
    For iLp = 0 To 11
        sChar = Mid$(sString, iLp, 1)
        Select Case iLp
            Case 0
                char1 = sChar
            Case 1
                char2 = sChar
            Case 2
                char3 = sChar
            Case 3
                char4 = sChar
            Case 4
                char5 = sChar
            Case 5
                char6 = sChar
            Case 6
                char7 = sChar
            Case 7
                char8 = sChar
            Case 8
                char9 = sChar
            Case 9
                char10 = sChar
            Case 10
                char11 = sChar
            Case 11
                char12 = sChar
        End Select
    Next iLp

03-01-2003, 08:50 PM

AcidGoat

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Hmmm... I tried that code and got this: Run-time error '5': Invalid procedure call or argument. Upon hitting debug, this line was highlighted: sChar = Mid$(sString, iLp, 1)

Maybe there's somethign I'm missing, but I've never seen code like this before (I haven't been using VB for very long, sorry).

03-01-2003, 09:16 PM

AcidGoat

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After a bit of tinkering and investigating I was able to get it to work. This is what I used:

sString = "goatgoatgoat"

char1 = Mid$(sString, 1, 1)
char2 = Mid$(sString, 2, 1)
char3 = Mid$(sString, 3, 1)
char4 = Mid$(sString, 4, 1)
char5 = Mid$(sString, 5, 1)
char6 = Mid$(sString, 6, 1)
char7 = Mid$(sString, 7, 1)
char8 = Mid$(sString, 8, 1)
char9 = Mid$(sString, 9, 1)
char10 = Mid$(sString, 10, 1)
char11 = Mid$(sString, 11, 1)
char12 = Mid$(sString, 12, 1)

03-01-2003, 09:16 PM

Kaluriel

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iLp could be 1 to 12. You can't start a character number 0

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#10

03-01-2003, 11:30 PM

diamond_panther diamond_panther is offline

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why do u want each letter to have its own variable?
isn't this EXACTLY what an array is for?

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