Retrieving the right-most portion of a string

General
1

Say I have the following strings:

[code]index.php/saint/elsewhere

and

/going/absolutely/nowhere[/code]

I need a function that will return:

elsewhere nowhere

I need to be able to grab all the characters that occur after the right-most ‘/’ character. If there is no ‘/’ character then I need to return all the characters.

Any idea of the most efficient way to do this?

2

Hi Gary,

look up the functions split, nthfield and countfields.

HTH

Trixi

4

[code] Dim SearchString as String = “kjs// sdfkl .22w#$4/trailing”
Dim SearchCharacter As UInt8 = Asc("/")

Dim ByteLength As Int32 = SearchString.LenB()
Dim MemBlock As New MemoryBlock(ByteLength)
MemBlock.StringValue(0, SearchString.Len()) = SearchString

Dim CurrentCharacter As UInt8
Dim Position As Int32 = 0

For i As Integer = ByteLength - 1 DownTo 0

CurrentCharacter = MemBlock.UInt8Value(i)

If (CurrentCharacter = SearchCharacter) Then
  
  Position = i + 2
  Exit
  
End If

Next

Dim TrailingString As String = SearchString.Mid(Position)

MsgBox Str(TrailingString)[/code]

You’re lucky I felt like a quick challenge this morning… :slight_smile:

EDIT: I’m pretty certain that this would fail, however, if your SearchString has characters with code values over 255 (or possibly, 127). Just something to be aware of, but I doubt this would impact what appears to be your intended usage.

5

trailing = url.NthField("/", url.CountFields("/"))

6

If you’re going to do this repeatedly, I’d see if a regular expression would be faster. One pattern would be:

[^/]+$

7

Thank you everybody.

[quote=17507:@Kem Tekinay]If you’re going to do this repeatedly, I’d see if a regular expression would be faster. One pattern would be:

[^/]+$[/quote]
Perfect.

This also works but is slightly slower than the RegEx.

8

One more thought:

dim parts() as string = url.Split( "/" )
dim trailing as string = parts( parts.Ubound )