EXCEL VBA - Delete if a column exists - vba

Question

I have a requirement wherein, I need to delete that Particular column if that exists.
I am trying to locate the Particular column through column header.
This is my code,
If sColumnName = (WorksheetFunction.Match("RSD", Sheets("RS_Report").Rows(1), 0)) And sColumnName = True Then
DDC= WorksheetFunction.Match("RSD", Sheets("RS_Report").Rows(1), 0)
DFC= GetColumnLetter(DDC)
Range(DFC& 1 & ":" & DFC& lastrow).Select
Selection.Delete Shift:=xlUp
the GetColumnLetter and lastrow are my user defined functions and they return correct values. I am not sure how to check if a column exists or not. Kindly help me with this. Share your thoughts.

Answer 1

you can simply go like this
Dim col As Variant
With Sheets("RS_Report") '<--| reference relevant worksheet
col = WorksheetFunction.Match("RSD", .Rows(1), 0) '<--| try searching its row 1 for wanted header text
If Not IsError(col) Then .Columns(col).Delete '<--| if found then delete its entire column
End With

Answer 2

There are three ways to do this.
1) A for loop that looks to the extents of the header row for a specific string.
Pro: its easy
Con: the string has to be exact
something like
Dim string as yourString
Dim lColumn As Long
lColumn = ws.UsedRange.Columns.Count
yourString = whatever
for x = 1 to lcolumn
if range(cells(1, 1), Cells(1, x)).Value = yourString
Columns(x).EntireColumn.Delete
End If
next
2) use the Range.Find method which you can learn about here https://msdn.microsoft.com/en-us/library/office/ff839746.aspx
Here is a short rough example you can use as reference:
Sub Header_Format()
Dim rLastCell As Range
Set rLastCell = UpdateSheet.Cells.Find(What:="*", After:=UpdateSheet.Cells(1, 1), LookIn:=xlFormulas, LookAt:= _
xlPart, SearchOrder:=xlByColumns, SearchDirection:=xlPrevious, MatchCase:=False)
With UpdateSheet.Range(Cells(4, 1), Cells(4, rLastCell.Column))
.Copy MasterSheet.Cells(1, 1)
.Copy RemovalSheet.Cells(1, 1)
End With
End Sub
3) Finally there is using the match method, which someone spoke on already.
https://msdn.microsoft.com/en-us/library/office/ff835873.aspx