call sub(arg1 as Range): Error 424: object required

Hans Troost

New Member
Joined
Jan 6, 2015
Messages
35
Office Version
  1. 365
  2. 2019
  3. 2016
Platform
  1. Windows
dear all

My question is only technical: how to call a sub expecting a Range argument, using a range.
For functionality the "poor man's solution" shown below works, so I can continue, but don't like it.

I have to call mysub (arg1 as Range) for a column in a table.

the "poor man's solution" that perfectly works is:

mysub(ActiveSheet.Range("table name[column name]")) ' as in ActiveSheet.Range("Uitgiftelijst[VB.nu]"))

But I don't find it very elegant, because I copy/pasted the table from another sheet.
So I have a named range "table name"
- and defined it as ListObject
- and did a lot of things (formatting etc.) with it (except calling a this sub)
- e.g. mytable.ListColumns("column name").DataBodyRange.Font.name = "font name" etc. etc.


dim myTable as Listobject
set myTable = activesheet.listobjects("table name")

But calling the mysub(arg1 as Range) fails all the time with error 424: Object required
What I tried is
mysub(myTable.ListColums("column name").DataBodyRange
mysub(myTable.ListColums("column name").Range
mysub(myTable.ListColums("column name") resulting in Compile Error, Data type mismatch


Question:
how to call a mysub (arg1 as Range) with the elegant use of my already defined table references?

Example table and code : see the code for what i tired to achieve. Kept the working version and commented out the error triggering lines:

CategoryJanFebMarAprMayJunTotal Column
Alpha533327495039251
Beta427841207462317
Charlie392164222741214
Deltal593331614537266
Echo293129554469257
Foxtrot433035446354269
Golf224335536572290
Hotel714241796131325


VBA Code:
Option Explicit

Sub CondForm()

Dim myTbl As ListObject

Set myTbl = ActiveSheet.ListObjects("myTable")

With myTbl.ListColumns("Category").DataBodyRange
    .FormatConditions.Add Type:=xlExpression, Formula1:="=$H2>260"
    .FormatConditions(.FormatConditions.Count).SetFirstPriority
    With .FormatConditions(1).Interior
        .PatternColorIndex = xlAutomatic
        .Color = vbRed
        .TintAndShade = 0.875
    End With
    .FormatConditions(1).StopIfTrue = False
End With

RemoveCF (ActiveSheet.Range("myTable[Category]")) ' Works! but prefer to not use Activesheet.<something> but functionality of my defined table: myTbl.ListColumns("Category")

'RemoveCF (myTbl.ListColumns("Category").DataBodyRange) ' Runtime error 424: Object Required. Why: this is a Range isn't it ?
'RemoveCF (myTbl.ListColumns("Category").Range) ' Runtime error 424: Object Required. Why: this too is a Range isn't it ?
'RemoveCF (myTbl.ListColumns("Category")) 'Compile error, type mismatch: understandable, is not a Range
'RemoveCF (myTbl.ListColumns("Category").Range.Select) ' Runtime error 424: Object Required
'what else?

End Sub

Sub RemoveCF(ByRef mySel As Range)

    Dim myCell As Range
    
    For Each myCell In mySel
        myCell.Interior.Color = myCell.DisplayFormat.Interior.Color
    Next myCell
    
    mySel.FormatConditions.Delete

End Sub

Any help appreciated

Kind regards, Hans Troost
 

Excel Facts

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How about
VBA Code:
RemoveCF myTbl.ListColumns("Category").DataBodyRange
 
Upvote 0
Solution
How about
VBA Code:
RemoveCF myTbl.ListColumns("Category").DataBodyRange
Thank Fluff, this works.

And of course, I should have tried and I don't know why: kind of a road block in my brain?

But besides that: what is the reason this works: is this that - with the parentheses - it is calling a function in stead of a sub what was the intention?

Anyway: I'm very happy with your answer. It leaded me out of my tunnel vision when looking for solutions.

Best regards, Hans
 
Upvote 0
You would need the brackets if you "called" the sub like
VBA Code:
Call RemoveCF(myTbl.ListColumns("Category").DataBodyRange)
but not the way you did it.
 
Upvote 0
Thanks again Fluff, I understand it now and I will never forget: my memory is better than my “blocked” brain. Hans
 
Upvote 0
You're welcome & thanks for the feedback.
 
Upvote 0

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