Vbscript to Extract text within a String

Hello,
I need vbscript code to extract data within a string.  Below are 2 examples of the type of string:

[onestring]twostring:[/onestring]2/40
[onestring]twostring:[/onestring]XY4/2

Need code that will go to the last occurrence of “/” and extract the text between it and the last occurrence of “]”.  In the two examples above, 2 and XY4 would be the output.

Need code that will go to the last occurrence of “/” and extract the text after it.  In the two examples above, 40 and 2 would be the output.

I’m sure this can be accomplished with text functions (mid, left, right, instr), but I am not sure of the parameters in each statement.  

Any help would be greatly appreciated.

Thanks
Diammond
DiammondAsked:
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Bill PrewCommented:
This should extract that text from a string called strTemp.

Mid(strTemp, instrrev(strTemp, "/")+1)

~bp
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vb_elmarCommented:
myString = "[onestring]twostring:[/onestring]2/40"
a = Split(myString, "/")
ub = UBound(a) - 1
result = a(ub)
cnt = InStr(result, "]") + 1
result = Mid(result, cnt)
MsgBox result
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wdosanjosCommented:
Here is another alternative:

s = "[onestring]twostring:[/onestring]2/40"

i = InStrRev(s, "/")
j = InStrRev(s, "]")

v1 = mid(s, j + 1, i - j - 1)
v2 = mid(s, i + 1)

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vb_elmarCommented:
myString = "[onestring]twostring:[/onestring]2/40"
a = Split(myString, "/")
ub = UBound(a) - 1
cnt = InStr(a(ub), "]") + 1
result = Mid(a(ub), cnt)
MsgBox result

Open in new window

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vb_elmarCommented:
Here is a function.

You can call it with ...
MsgBox myFunction("[onestring]twostring:[/onestring]XY4/2")

Function myFunction(ByVal L As String) As String
    a = Split(L, "/")
    ub = UBound(a) - 1
    cnt = InStr(a(ub), "]") + 1
    myFunction = Mid(a(ub), cnt)
End Function

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vb_elmarCommented:
Here is a short function. You can call it with ...

tmp = "[onestring]twostring:[/onestring]XY4/2"
a = myFunction(tmp)
MsgBox a

Function myFunction(ByVal L As String) As String
  L1 = Mid(L, InStrRev(L, "]") + 1)
  myFunction = Mid(L1, 1, InStr(1, L1, "/") - 1)
End Function

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DiammondAuthor Commented:
All,

Thank you for responding.
The answer given by wdosanjos fits my need best.

Diammond
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