Solved

How to parse out the string into three strings in Visual basic

Posted on 2002-03-22
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Last Modified: 2013-11-25
I got a problem like, i got a string like:

InText = "TEST RETAIL 00001 Default"

I want the result like this:

L = TEST RETAIL
N= 00001
R = Default

By the below procedure i'm getting the result like:

L = TEST RETAIL
N = 1
R = 000 Default

Public Sub TrippleAtNumber(InText As String, L As String, N As String, R As String)
   Dim i As Long
   For i = 1 To Len(InText)
   'If Val(Mid(InText, i)) > 0 Or Mid(InText, i, 1) = "0" Then

       If Val(Mid(InText, i)) > "0" Or Mid(InText, i, 1) = "0" Then
           L = Trim(Left(InText, i))
           N = Val(Mid(InText, i))
           R = Trim(Mid(InText, i + Len(Str(Val(Mid(InText, i))))))
           Exit Sub
       End If
   Next
End Sub


It works fine for other strings but only if in the middle
starts with 0(zero) it gives problem. How to get rid of the problem.

VM
0
Comment
Question by:vmandem
  • 5
  • 4
  • 2
  • +2
14 Comments
 
LVL 20

Expert Comment

by:hes
ID: 6890029
Try this

Dim TempArr

TempArr = Split(InText," ")
L = TempArr(0) & " " & TempArr(1)
N = TempArr(2)
R = TempArr(3)
0
 

Author Comment

by:vmandem
ID: 6890039
Hes
I'm using VB 5.0 and the reason i'm using the above procedure is because in my strings i will get more spaces in the middle of the string but i'm splitting the string based on the number in the middle.

I consider the left string untill i see the number in the middle.

The procedure works fine for everything but not if the number starts with zero.

Like i said: Test Retail 00001 Default

It works fine when it is: Test Retail 10001 Default

gives, L = Test Retail
       N = 10001
       R = Default

VM

Please correct my procedure i appreciate your response
0
 
LVL 20

Expert Comment

by:hes
ID: 6890048
Try changing
If Val(Mid(InText, i)) > "0" Or Mid(InText, i, 1) = "0" Then

to

If Val(Mid(InText, i)) >= 0 Then
0
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Author Comment

by:vmandem
ID: 6890061
hes

Then i get

L = "T"
N = "0"
R = "EST 00001 DEFAULT"

How to resolve this

I appreciate your response.

Thanks
VM
0
 
LVL 44

Accepted Solution

by:
bruintje earned 15 total points
ID: 6890064
and with

Public Sub TrippleAtNumber()
  Dim i As Long
 
  InText = "TEST RETAIL 00001 Default"


  For i = 1 To Len(InText)
  'If Val(Mid(InText, i)) > 0 Or Mid(InText, i, 1) = "0" Then

      If Val(Mid(InText, i)) > "0" Or Mid(InText, i, 1) = "0" Then
          L = Trim(Left(InText, i))
          pos = InStr(1, (Mid(InText, i + 1)), " ", vbTextCompare)
          N = Trim(Mid(InText, i, pos))
          R = Trim(Right(InText, Len(InText) - i - pos))
          Exit Sub
      End If
  Next
End Sub
0
 
LVL 20

Expert Comment

by:hes
ID: 6890093
Try

For i = 1 To Len(InText)
  'If Val(Mid(InText, i)) > 0 Or Mid(InText, i, 1) = "0" Then
    If IsNumeric(Mid(InText, i, 1)) Then
      If CInt(Mid(InText, i, 1)) >= 0 And CInt(Mid(InText, i, 1)) <= 9 Then
          L = Trim(Left(InText, i - 1))
          N = Mid(InText, i, 5)
          R = Trim(Mid(InText, i + 5))
          Exit Sub
      End If
    End If
  Next
0
 
LVL 20

Expert Comment

by:hes
ID: 6890170
Actually you can get rid of the second if

 For i = 1 To Len(InText)
   If IsNumeric(Mid(InText, i, 1)) Then
     L = Trim(Left(InText, i - 1))
     N = Mid(InText, i, 5)
     R = Trim(Mid(InText, i + 5))
     Exit Sub
   End If
  Next
0
 
LVL 6

Expert Comment

by:sharmon
ID: 6890496
If you don't know how long the spaces may be but you always have three portions seperated with a numeric value in the middle, this routine should work for you just fine.  I am not sure if you're data always follows that format but if so, try this out.


Option Explicit

Public Sub Test()
    Dim strTest As String
    Dim L As String
    Dim N As String
    Dim R As String
   
    strTest = "TEST RETAIL 00001 Default"
    TrippleAtNumber strTest, L, N, R

    Debug.Print L
    Debug.Print N
    Debug.Print R
End Sub

Public Sub TrippleAtNumber(InText As String, L As String, N As String, R As String)
    Dim i As Long
    Dim blnFoundNumeric As Boolean
    Dim strChar As String
   
    blnFoundNumeric = False
    L = ""
    N = ""
    R = ""
   
    For i = 1 To Len(InText)
        strChar = Mid$(InText, i, 1)
        If Not IsNumeric(strChar) Then
            If Not blnFoundNumeric Then
                L = L & strChar
            Else
                R = R & strChar
            End If
        Else
            N = N & strChar
            blnFoundNumeric = True
        End If
    Next
   
    L = Trim$(L)
    N = Trim$(N)
    R = Trim$(R)
End Sub
0
 
LVL 10

Expert Comment

by:smegghead
ID: 6891038
Try the following.

  Dim i As Long
  For i = 1 To Len(InText)
      If (Asc(Mid$(InText, i)) >= 48) And (Asc(Mid$(InText, i)) <= 57) Then ' i.e. it's a number
          L = Trim(Left(InText, i - 1))
          N = Mid(InText, i, InStr(i, InText, " ") - i)
          R = Trim(Mid(InText, i + Len(N)))
          ' N=CSTR(VAL(N)) ' Uncomment this line if you want the value to be truncated
      End If
  Next


I think the problem with your original routine was that you were converting the string into a number, which reduced it's size from '000065' down to '65' or whatever.
0
 
LVL 10

Expert Comment

by:smegghead
ID: 6891044
or even...

  For i = 1 To Len(InText)
      If Mid$(InText, i, 1) Like "#" Then
          L = Trim(Left(InText, i - 1))
          N = Mid(InText, i, InStr(i, InText, " ") - i)
          R = Trim(Mid(InText, i + Len(N)))
      End If
  Next
0
 

Author Comment

by:vmandem
ID: 6891180
bruintje

I accept your comment as answer.

Hes

I really appreciate your comments too. It is not that your comments are wrong, but i got solved my problem with bruintje comment.

I really appreciate every one's comment.

I thank all of them for there valuable comments.

I hope we will continue the same in future.

Thanks
VM
0
 
LVL 20

Expert Comment

by:hes
ID: 6891183
No problem, glad you got it solved
0
 
LVL 44

Expert Comment

by:bruintje
ID: 6891204
thanks for the points, have to admit that this thread is full of usefull solutions
0
 

Author Comment

by:vmandem
ID: 6894510
bruintje

I got a weird problem on the one we were discussing.

I hope you can give me an answer.

I got string like:

TEST RETAIL 00000 Default

If everything is zero's in the middle, it is considering
like this:
str1: TEST RETAIL 0
str2:00000
str3:Default

Everthing is fine but in str1 it is showing zero at the end. I don't no why.

Could you please take a moment to look into that.

VM
0

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