Find InStr: (cant explain)

[|2eM!x]

I wrote this code, that will search in a string in between an opening and closing phrase:

Private Sub Form_Load()
    MsgBox FindInStr("||", "::", "Cat alpha Pensi france catarax || alabastar cheese gato faggle :: google cheese")
End Sub
 
Public Function FindInStr(StrOpening As String, StrClosing As String, lpString As String) As String
        Dim x As Integer, y As Integer, InStrStr As String, FinalStr As String
            x = InStr(lpString, StrOpening)
                x = x + Len(StrOpening)
               FinalStr = Mid(lpString, x, Len(lpString) - x)
                y = InStr(FinalStr, StrClosing)
                    InStrStr = FinalStr
                   FinalStr = Mid(FinalStr, y, Len(FinalStr) - y + 1)
                FindInStr = Replace(InStrStr, FinalStr, vbNullString): FindInStr = Trim(FindInStr)
End Function

This will return "alabastar cheese gato faggle"

This also didnt work as i wanted, i didnt want to use replace, but i guess i had to

[szlamany]

This seems to do the same things without the REPLACE...

Why did you want the REPLACE?

Private Sub Form_Load()
    MsgBox FindInStr("||", "::", "Cat alpha Pensi france catarax || alabastar cheese gato faggle :: google cheese")
End Sub

Public Function FindInStr(StrOpening As String, StrClosing As String, lpString As String) As String
        Dim x As Integer, y As Integer, InStrStr As String, FinalStr As String
            x = InStr(lpString, StrOpening)
            x = x + Len(StrOpening)
            y = InStr(x, lpString, StrClosing)
            FindInStr = Trim(Mid(lpString, x, (y - x) - 1))
End Function

[|2eM!x]

i didnt want to use replace

i couldnt figure out how this last bit:
FindInStr = Trim(Mid(lpString, x, (y - x) - 1))

[CVMichael]

Check out the LeftRange function I posted in this thread:
VB - Some functions that makes string parsing easier
I use thouse functions in 99% of my applications, very usefull functions...