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Loop through a row and delete all columns when a certain value is found

Posted on 2013-01-10
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Last Modified: 2013-01-28
Hi Experts

I need to be able to loop through the column headings in Row 1 of a spreadsheet (starting from column C  i.e "C1" and find the first column which contains value "Old_Address" and then delete this columns and all columns  after this.  So for example if "Old_Address" was found in Cell H1 then I would want all the columns from H onwards to be deleted.  The number of columns will change everyday.

Your assistance would be appreciated.

Thanks
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Comment
Question by:simsima_7876
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9 Comments
 
LVL 34

Expert Comment

by:Norie
ID: 38764192
Try this.
Dim varOldAdd As Variant
Dim LastCol As Long

    varOldAdd = Application.Match("Old_Address", Rows(1), 0)
    LastCol = Cells(1, Columns.Count).End(xlToLeft).Column

    If Not IsError(varOldAdd) Then
        Cells(1, varOldAdd).Resize(, LastCol - varOldAdd + 1).EntireColumn.Delete
    End If

Open in new window

0
 

Author Comment

by:simsima_7876
ID: 38764243
I get a error 2042 with this code even though the value is definitely there.
0
 
LVL 26

Expert Comment

by:redmondb
ID: 38764267
Hi,  simsima_7876.

Please see attached. The code is...
Option Explicit

Sub Delete_Columns()
Dim xCol      As Range
Dim xLast_Col As Long
Dim i         As Long

Sheets("To Be Deleted").Activate

If ActiveSheet.UsedRange.Columns.Count > 0 Then xLast_Col = [A1].SpecialCells(xlLastCell).Column
If xLast_Col < 4 Then
    MsgBox ("No columns after C - run cancelled.")
    Exit Sub
End If

Set xCol = Range("1:1").Find(What:="Old_Address", LookIn:=xlValues, LookAt:=xlWhole, MatchCase:=False)
If xCol Is Nothing Then
    MsgBox ("""Old_Address"" not found.")
Else
    Range(Cells(1, xCol.Column), Cells(1, xLast_Col)).EntireColumn.Delete
    If ActiveSheet.UsedRange.Columns.Count < 0 Then Debug.Print "?!"
End If

End Sub

Open in new window

Regards,
Brian.
Delete-Columns.xls
0
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Author Comment

by:simsima_7876
ID: 38764269
However in this case the value I am searching for is 41723 ( a numeric date) as opposed to the string Old_address
0
 
LVL 26

Expert Comment

by:redmondb
ID: 38764289
Apologies, simsima_7876, I forgot to skip columns A:B. Please see attached.Delete-Columns-V2.xls
0
 
LVL 34

Expert Comment

by:Norie
ID: 38764314
simsima

Did you change the code I posted in anyway?
0
 
LVL 26

Accepted Solution

by:
redmondb earned 1600 total points
ID: 38764341
simsima,

The attached handles dates. The code is...
Option Explicit

Sub Delete_Columns()
Dim xCol      As Range
Dim xLast_Col As Long
Dim i         As Long
Dim xSearch   As Variant

Sheets("To Be Deleted").Activate

If ActiveSheet.UsedRange.Columns.Count > 0 Then xLast_Col = [A1].SpecialCells(xlLastCell).Column
If xLast_Col < 3 Then
    MsgBox ("No columns after B - run cancelled.")
    Exit Sub
End If

xSearch = DateSerial(2014, 3, 25)

Set xCol = Range(Cells(1, 3), Cells(1, xLast_Col)).Find(What:=xSearch, LookIn:=xlFormulas, LookAt:=xlWhole, MatchCase:=False)
If xCol Is Nothing Then
    MsgBox (xSearch & " not found.")
Else
    Range(Cells(1, xCol.Column), Cells(1, xLast_Col)).EntireColumn.Delete
    If ActiveSheet.UsedRange.Columns.Count < 0 Then Debug.Print "?!"
End If

End Sub

Open in new window

Regards,
Brian.Delete-Columns-V3.xls
0
 

Author Closing Comment

by:simsima_7876
ID: 38826970
Hi Redmond - sorry for the delay.  You arethe man!  Thanks for the solution - it worked like  a dream.
0
 
LVL 26

Expert Comment

by:redmondb
ID: 38827029
Thanks, simsima!
0

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