Removing the trailing comma from a string

[KashanKhan]

hi there

i have this string [code=vb]
dim str as string

str="1,2,3,4,"[/code]

i required this
[code=vb]
str="1,2,3,4"[/code]

can anyone help..!

[pbmods]

Heya, Kashan. Welcome to TSDN!

Please use CODE tags when posting source code:

[CODE=vb]
Visual Basic source code goes here.
[/CODE]

I'm going to go ahead and move this thread to the VB forum, where our resident Experts will be better able to help you out.

[Killer42]

i required this
[code=vb]
str="1,2,3,4"[/code]

In what version of VB?

[jamesd0142]

In what version of VB?

There is a substring method in VB.Net, or a mid function in VB.

Either one of these will allow you to pick out characters at set positions and allow you to delete them.

You could for example in VB.
[code=vb]
Dim str As String = "1,2,3,4,"
Dim str1 As String

Dim i As Integer = Len(str)
str1 = Mid(str, 1, i - 1)
[/code]

This would pick out all characters except the last one which will be your comma, and store in str1.

Thanks JAmes

[Killer42]

Thanks for that, James.

However, if you're going to post code, please try to ensure it's working code.

Dim str As String = "1,2,3,4,"
...is VB.Net syntax and won't work in VB6 or earlier. In those versions you would need to do something like...

[CODE=vb]Dim str As String
str = "1,2,3,4,"[/CODE]

[jamesd0142]

ahh i see, well thats a new 1 to me :P

I only use vb.net and didnt realise the small differences.

[Killer42]

ahh i see, well thats a new 1 to me :P

I only use vb.net and didnt realise the small differences.

:)
That's probably just about the smallest of the differences. Some are huge.

I downloaded VB 2005 Express Edition weeks ago. But I haven't done anything with it yet because I just don't know where to start. There's so much to learn.

[jamesd0142]

yes, maybe i should add that at home i use vb.net, and in work i downloaded the free vb 2005 express edition also, so im having to switch between the two! annoying!

although when i finally tried to use this i find little differences.

for example:

dim a as string = "123" works for both.

I understand this does'nt work for all versions of vb as you previously told me.

some versions require this???

dim a as string
a = "123"

[ghendric1]

VB6 functions to trim characters

I know this post may be pretty old but I found myself working in old VB6 code once more and had to create a couple of functions that removes characters from the start and end of a string so I wrote these functions. Maybe someone will find these useful if they're working with VB6 again.

Public Function TrimEnd(ByVal someString As String, ByVal someChar As String) As String

    On Error GoTo Err_Proc
    someString = Trim(someString)
    someChar = Trim(someChar)
    
    If Len(someString) = 0 Then GoTo Exit_Proc
    If Right(someString, Len(someChar)) <> someChar Then GoTo Exit_Proc

    While Right(someString, Len(someChar)) = someChar
        someString = Left(someString, Len(someString) - Len(someChar))
    Wend

Exit_Proc:
    TrimEnd = someString
    Exit Function

Err_Proc:

    'Err_Handler True, Err.Number, Err.Description, Erl, "Form1", "Function TrimEnd"
    MsgBox Err.Description
    Err.Clear
    Resume Exit_Proc

End Function

Public Function TrimStart(ByVal someString As String, ByVal someChar As String) As String

    On Error GoTo Err_Proc
    someString = Trim(someString)
    someChar = Trim(someChar)

    If Len(someString) = 0 Then GoTo Exit_Proc
    If Left(someString, Len(someChar)) <> someChar Then GoTo Exit_Proc

    While Left(someString, Len(someChar)) = someChar
        someString = Mid(someString, Len(someChar) + 1)
    Wend

Exit_Proc:
    TrimStart = someString
    Exit Function

Err_Proc:

    'Err_Handler True, Err.Number, Err.Description, Erl, "Form1", "Function TrimStart"
    MsgBox Err.Description
    Err.Clear
    Resume Exit_Proc
End Function