In the process of responding to a post in another forum, I came across this function, which I hadn't used in years.
It allows the user to specify a delimiter and designated block (e.g. 1,2,3) and returns that block.
I don't know if I wrote it (would like to think so) or copied it from an unknown source. A search failed to turn up the function name. If it's yours, please let us know.
It eliminates a bunch of gyrations trying to get just that specific block of characters.
HTH - Bob
It allows the user to specify a delimiter and designated block (e.g. 1,2,3) and returns that block.
I don't know if I wrote it (would like to think so) or copied it from an unknown source. A search failed to turn up the function name. If it's yours, please let us know.
Code:
Public Function PassbackAnyword(pstrText As String, pintword As Integer, pstrdivider As String) As Variant
'Purpose: Given a string, a word and the divider,
' returns the specific word.
'Input: ? PassbackAnyword("1234*7890*8888", 2, "*")
'Output: 7890
Dim intLoop As Integer
Dim intPos As Integer
Dim intprev As Integer
Dim varstring As Variant
'Don't waste your time if the divider isn't in the string
If InStr(pstrText, pstrdivider) <> 0 Then
intPos = 1
intprev = 1
pstrText = pstrText & pstrdivider
For intLoop = 1 To pintword
intPos = InStr(intprev + 1, pstrText, pstrdivider)
If intPos <> 0 Then
If intLoop < pintword Then
intprev = intPos
End If
Else
intPos = intprev
End If
Next
varstring = Mid(pstrText, intprev, intPos - intprev)
If pintword > 1 And varstring <> "" Then
varstring = Right(varstring, Len(varstring) - 1)
End If
Else
'If it's the first word we want then it's all the string otherwise is nothing
If pintword = 1 Then
varstring = pstrText
End If
End If
If varstring = "" Or varstring = pstrdivider Then
varstring = Null
End If
PassbackAnyword = varstring
End FunctionIt eliminates a bunch of gyrations trying to get just that specific block of characters.
HTH - Bob
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