InStr 提取“[”之前的所有文本，如果某些单元格包含“[”而某些单元格不包含

Question

I have the following piece of code that returns a

Run-time error '5'

as soon as it comes across a cell that do not contain [

Cells(i + 2, 11) = Left(Cells(i * 2 + 1, 12), InStr(1, Cells(i * 2 + 1, 12), "[") - 1)

In this example:
Cheese[1] is ok and returns Cheese .
Marmite(2)[13] is ok and returns Marmite(2) .
Spreadable butter(13) doesn't work.

What is the fix please

---

Thank you, so following on from the help below how do I incorporate and apply two IF statements into the following?

For i = 1 To 4
   If Cells(i * 2 + 2, 14) <> "" Then
      Cells(i + 2, 3) = Left(Right(GetURL(Cells(i * 2 + 2, 17)), 12), 7)
      Cells(i + 2, 11) = Left(Cells(i * 2 + 1, 12), InStr(1, Cells(i * 2 + 1, 12), "[") - 1)
      Cells(i + 2, 12) = Left(Cells(i * 2 + 1, 13), InStr(1, Cells(i * 2 + 1, 13), "[") - 1)
   End If
Next i

I am only a beginner at VBA.

Answer 1

You can just check to see if it contains a [ first...

The Instr() function returns a 0 if the string you are looking for cannot be found, so without an If statement to check for that first, you end up with a 0, and your code is subtracting 1 from the zero to get -1, and that is not a valid argument for the Left function, so you get the error.

If InStr(1, Cells(i * 2 + 1, 12), "[") > -1 Then
  Cells(i + 2, 11) = Left(Cells(i * 2 + 1, 12), InStr(1, Cells(i * 2 + 1, 12), "[") - 1)
End If

This assumes the cell will never actually start with a [

To avoid getting an error if it IS the first character:

If InStr(1, Cells(i * 2 + 1, 12), "[") > 0 Then
  Cells(i + 2, 11) = Left(Cells(i * 2 + 1, 12), InStr(1, Cells(i * 2 + 1, 12), "[") - 1)
End If