Range.Find() text with carriage return Excel VBA - vba

Question

What I'm trying to do
Locate the column whose header cell contains a unique string. In other words, I know the cell's text, and I know the cell is in row 1, but I don't know which column. NOTE: I want to search for the entire text, not just part of it. NOTE2: The text can vary, so I cannot hardcode the value into my code. Rather I need to use the variable in which the value is stored.
The problem
When there's no carriage return in the header text, a simple newCol = Range("1:1").Find(headerText).Column works fine. However, if there is a carriage return, this doesn't work. It throws up the error "Object variable or With block variable not set". Here's my exact header string:
Incomplete Email
(more text)
What I've already tried
I also tried using WorksheetFunction.Match(headerText, Range("1:1"), 0), but got the same issue.
Additional notes and requirements
This is part of an add-in, so I do not want to change anything in the user's excel sheet if I don't have to (i.e., I don't want to remove the carriage return).
Technically, I'm doing this in a function:
Public Function getColumn(headerText As Variant)
getColumn = Range("1:1").Find(headerText).Column
End Function
Thanks!

Answer 1

pls try with below code
Public Function getColumn(headerText As String)
str1 = Split(headerText, vbCrLf)
str2 = UBound(str1)
b = Range("1:1").Find(str1(0) & Chr(10) & str1(1)).Column
End Function

Answer 2

Here's the thing: text with and without line break is NOT the same text hence the .Find fail. What you should do is a pattern lookup. I have just tested this and it works, provided that if there is no line break there shall be a space:
Sub test()
Dim rex As RegExp, ran As Range
Dim col As Integer, headerText As String
'read you headerText here
Set rex = New RegExp
rex.Pattern = RegexIt(headerText)
For Each ran In Range("1:1")
If rex.test(ran.Text) Then
col = ran.Column
Exit For
End If
Next ran
MsgBox col
End Sub
Function RegexIt(what As String) As String
what = Replace(what, "(", "\(")
what = Replace(what, ")", "\)")
what = Replace(what, "[", "\[")
what = Replace(what, "]", "\]")
what = Replace(what, "<", "\<")
what = Replace(what, ">", "\>")
what = Replace(what, " ", "[\n ]?")
what = Replace(what, vbCrLf, "[\n ]?")
End Function
Good luck!
Edit: Reference to Microsoft VBScript Regular Expressions 5.5 required
Edit2: Edited for variable use. Explanation: Replace space in variable value with optionel space/line break, escape brackets for pattern matching.

Answer 3

Your code should work even if the header cell contains carriage returns:
Sub FindColumnWithTextInRowOne()
Dim headerText As String, newCol As Long
headerText = "whatever"
newCol = Range("1:1").Find(headerText).Column
MsgBox newCol
End Sub
This is because your use of Find() does not require a match to the WHOLE contents of the cell.
EDIT#1:
If the header cell was constructed using a formula, then a slightly different Find() should be used:
Sub FindColumnWithTextInRowOne()
Dim headerText As String, newCol As Long, r As Range
headerText = Range("H1").Text
newCol = Range("1:1").Find(What:=headerText, LookAt:=xlWhole, LookIn:=xlValues).Column
MsgBox newCol
End Sub