vbworkingwithstrings.txt

字符串处理函数.

--7-------------------------------------------------------------
Date sent:        Fri, 16 May 1997 13:36:53 -0500 (CDT)
From:             Tarik <[e-mail]>

Suppose I copy a data from the excel to the clipboard, which is as
follows:

Allen	12
Anderson 13
Douglas 12
Ohio 49


How can I read the county name and the data in an array or in a grid box?
the Clipboard.GetText(vbCFText) reads the whole data, i want to read one
by one.

make a new project, a form, two labels and a commandbutton

Private Sub Command1_Click()
    Dim vTekst$
    
    vTekst$ = "Allen 12 "
    vTekst$ = vTekst$ & "Anderson 13 "
    vTekst$ = vTekst$ & "Bernard 14 "
    vTekst$ = vTekst$ & "Constance 15 "
    Label1.Caption = vTekst$
    
    Select Case Command1.Caption
    Case "Copy Clipboard"
        Clipboard.Clear
        Clipboard.SetText Label1.Caption
        Command1.Caption = "Put into Label"
    Case "Put into Label"
        Label2.Caption = GetPartofString(Clipboard.GetText, 1)
        Command1.Caption = "Copy Clipboard"

        'read in array
        Dim vText(7) As String
        Dim c%
    
        For c% = 0 To 7
            vText(c%) = GetPartofString(Clipboard.GetText, c% + 1)
        Next c%
        
        'show result
        For c% = 0 To 7
            MsgBox vText(c%)
        Next c%

End Sub

Private Function GetPartofString(source$, part%) As String
    Dim p%, c%, tmp$
    
    tmp$ = source$
    c% = 0
    Do
        p% = InStr(tmp, Chr(32))
        If p% <> 0 Then
            GetPartofString = Left(tmp, p% - 1)
            c% = c% + 1
            tmp = Right(tmp, Len(tmp) - p%)
        End If
    Loop While c% <> part%
    
End Function


--6-------------------------------------------------------------

On 12 May 1997 11:57:58 GMT, "Aviv Shaham" <[e-mail]> wrote:

>hi
>
>How can I count the charaters in a string
>For example:
>I want to count the characters in Text1.Text when a use presses OK.
>
How about the command LEN(Text1.text)? You get the lenght.. 
including spaces and other non-charaters. 
So if you want to loose then too you must use a litte function to check 
if a character of text1.text is really a text-charater:



Public Function CountCharacters(source$) As Integer
    Dim counter%, t%
    Const Characters$ = "abcdefghijklmnopqrstuvwxyz"

    For t% = 1 To Len(source$)
        If InStr(Characters, LCase$(Mid$(source$, t%, 1))) <> 0 Then
            counter% = counter% + 1
        End If
    Next t%
    CountCharacters = counter%
End Function



use as

vString$ = "Testing .... about what?"
MsgBox CountCharacters(vString$)

--5-------------------------------------------------------------

On 9 May 1997 02:36:50 GMT, [e-mail] (XX007BOND) wrote:

>Does anyone know the code needed to make a special kind of loop.  I need
>one that as an example takes the string "Hey how are you" and places each
>letter in a different variable that increases on its one. Ex.  H goes in
>A1. e goes in A2, y goes in A4 and so on.  Thanx for your help

Dim vChar() as String

Sub PlaceInArray(source$)

Dim p%

	For p% = 1 To Len(source$)
		Redim Preserve vChar(p%)
		vChar(p%) = Mid$(source$,p%,1)
	Next p%

End Sub


In the array vChar are all the characters of the given key. 
You can use them by calling the right item: ergo Msgbox vChar(1)



--4-------------------------------------------------------------
On Tue, 06 May 1997 10:06:29 -0600, Oyvind Finnesand <[e-mail]> wrote:

>How do you read words one by one from a text file into string variables? 
>Each word in the text file is separated by a single space.


Dim vWords() as String

Sub SplitStringintoWords(bron$)
Dim c%, p%, t%, vCheck%
Dim TempBron$, tmp$

'splitting the input into words
    t% = 0
    TempBron$ = bron$
    For c% = 1 To Len(bron$)
        p% = InStr(TempBron$, Chr(32))
        If p% <> 0 Then
        ReDim Preserve vWords(t%)
            tmp = Left$(TempBron$, p% - 1)
            vWords(t%) = StripString(tmp)
            TempBron$ = Right$(TempBron$, Len(TempBron$) - p)
            t% = t% + 1
            c% = c% + p%
        End If
    Next c%
    ReDim Preserve vWords(t%)
    vWords(t%) = StripString(TempBron)

End Sub


First you must read in one line of the textfile and then use the above proc. 
In the array vWords are all the words from the textfiles.
If you need help reading the file let me know

--3-------------------------------------------------------------
On 4 May 1997 21:52:41 GMT, in comp.lang.basic.visual.misc you wrote:

>
>I need to know how to replace the following text with the text next to it
>on all my text boxes on a form and at run time.  Thanks ahead of time.
>

try this...
BTW it comes form two procedures on the page below..

'make a new form with a command button and 3 textboxen
'add the code below
'press F5
'and then the command button

Private Sub Command1_Click()
    Const vString$ = "testing for somebody on the net"
    Select Case Command1.Caption
    Case "&Replace text in all textboxes"
        Text1.Text = vString
        Text2.Text = vString
        Text3.Text = vString
        Command1.Caption = "&Replace somebody with me"
    Case "&Replace somebody with me"
        Call ChangeText
        Command1.Caption = "&Replace text in all textboxes"
    End Select
    
End Sub

Function sReplace(SearchLine As String, SearchFor As String, _ ReplaceWith As String)
    Dim vSearchLine As String, found As Integer
    
    found = InStr(SearchLine, SearchFor): vSearchLine = SearchLine
    If found <> 0 Then
        vSearchLine = ""
        If found > 1 Then vSearchLine = Left(SearchLine, found - 1) _
        vSearchLine = vSearchLine + ReplaceWith
        If found + Len(SearchFor) - 1 < Len(SearchLine) Then _
            vSearchLine = vSearchLine + Right$(SearchLine, _ 
	    Len(SearchLine) - found - Len(SearchFor) + 1)
    End If
    sReplace = vSearchLine
    
End Function

Private Sub ChangeText()
    Dim Control

    For Each Control In Form1.Controls
        If TypeOf Control Is TextBox Then
            Control.Text = sReplace(Control.Text, "somebody", "me")
        End If
    Next Control
        
End Sub


--2-------------------------------------------------------------

On Tue, 29 Apr 1997 22:19:46 -0300, in comp.lang.basic.visual.misc you wrote:

>I have a text box so I (as user) can enter a string.  I have a control
>array of 60 little text boxes with words in them for my dictionary.  I
>am trying to check each word in my string to see if it matches a word in
>my array.  I have to remove the puncuation before matching the word.  If
>the whole sentence is spelled ok I send out a message.  If not I send
>out a different message.
>
>Any general advice on how you would procede with this would be great.

Assuming you have a textbox where you enter one or more words and you want to check 
it against an array.. you can use the next code as a starting point..
But consider of making your spellingchecker list a database. When it is growing a 
database works much faster.


'on the general section
Dim vWords()
Dim Max%
Dim vCheckWord()

Private Form1_Load()
    Text1.SetFocus
    Text1.Text = "living in america but not really"
    Max% = 10
    
    'make array
    'you can use an ascii file to get the words
    Redim vCheckWord(Max)
    vCheckWord(0) = "walther"
    vCheckWord(1) = "musch"
    vCheckWord(2) = "america"
    vCheckWord(3) = "tilburg"
    vCheckWord(4) = "hallo"
    vCheckWord(5) = "testen"
    vCheckWord(6) = "testing"
    vCheckWord(7) = "really"
    vCheckWord(8) = "visual"
    vCheckWord(9) = "basic"

End Sub

Sub SplitStringintoWords(bron$)
Dim c%, p%, t%, vCheck%
Dim TempBron$, tmp$
Dim vOke As Boolean

'splitting the input into words
    t% = 0
    TempBron$ = bron$
    For c% = 1 To Len(bron$)
        p% = InStr(TempBron$, Chr(32))
        If p% <> 0 Then
        ReDim Preserve vWords(t%)
            tmp = Left$(TempBron$, p% - 1)
            vWords(t%) = StripString(tmp)
            TempBron$ = Right$(TempBron$, Len(TempBron$) - p)
            t% = t% + 1
            c% = c% + p%
        End If
    Next c%
    ReDim Preserve vWords(t%)
    vWords(t%) = StripString(TempBron)

'checking against spellingschecker
    vOke = False
    For c% = 0 To t%
        For vCheck% = 0 To Max
            If vCheckWord(vCheck%) <> vWords(c%) Then
                vOke = False
            Else
                vOke = True
                vCheck% = Max%
            End If
        Next vCheck%
        If Not vOke Then MsgBox vWords(c%)
        vOke = False
    Next c%

End Sub


Private Sub Text1_KeyPress(KeyAscii As Integer)
    If KeyAscii = 13 Then
        'split string into words
        Call SplitStringintoWords(Text1.Text)
    End If
        
End Sub

Function StripString(source As String) As String
Const Letters$ = "abcdefghijklmnopqrstuvwxyz"
Dim p%, tmp$

    tmp = source$
    For p% = 1 To Len(source$)
        If InStr(Letters, LCase(Mid$(source$, p%, 1))) = 0 Then
            Select Case p%
            Case 1
                tmp = Right$(source$, Len(source$) - p%)
            Case Len(source$)
                tmp = Left$(source$, Len(source$) - 1)
            Case Else
                tmp = Left$(source$, p%) & Right$(source$, Len(source$) - p%)
            End Select
        End If
    Next p%
    StripString = tmp
            
End Function



--1-------------------------------------------------------------

On 14 Apr 97 22:13:57 GMT, [e-mail] (amsaa) wrote:

>Hi,
>
>I need help in determining if a filename has an extension and then stripping it 
>off. Basically I want to get filename and save it as a backup file with a new 
>extension. Is there an easy way to do this without searching for all "/" in the 
>entire path string? 
>

start at the back of the path string and search for the first "\"
take the right$; this will give you the filename
If you want to delete the extension then do the same but search for the first ".". 
Beware that a filename in w95 can have more then one points!

code like this:

Dim p%

Function GetFileName(PathString$) As String

GetFileName = PathString
For p% = Len(PathString) To 0 Step -1
	If Mid$(PathString,p%,1) = "\" then 
		GetFileName = Right$(PathString,p%)
		Exit Function
	End If
Next p%
End Function