=1*MID(D3,2+MATCH(FALSE,IS
As shown above, the formula returns a #VALUE! error if a non-numeric character follows the first digit.
=1*MID(D3,MATCH(FALSE,ISERROR(1*MID(D3,ROW($3:$12),1)),0),255)
Extract.xlsm
'Install code in a regular module sheet
Function GetNumber(sText As String, Optional iStart As Integer = 3, Optional iLast As Integer = 12)
Dim i As Integer
GetNumber = ""
For i = iStart To iLast
If IsNumeric(Mid(sText, i, 1)) Then
GetNumber = Val(Mid(sText, i))
Exit For
End If
Next
End Function
=1*MID(D3,MATCH(FALSE,ISERROR(1*MID(D3,ROW($1:$10),1)),0),255)
The only returned value is 123 for all cells. Does that mean I'll have to repeat the array formula for E3:E12?Title | # Comments | Views | Activity |
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