Sub that does a sort gives error "You cannot change part of an array"

OaklandJim

Well-known Member
Joined
Nov 29, 2018
Messages
833
Office Version
  1. 365
Platform
  1. Windows
I cannot figure out what to look at to fix this issue. Sub used for sorting worked well until it did not. Now I get the error message "You cannot change part of an array" on the .Apply method. Below is the offending sub and debug.print output.

What am I doing wrong? Syntax? It must be something simple because, like I said, it worked before. I don't know of any changes to the dataset that I have made.

VBA Code:
Sub Sort_Events_Master()

    Dim rAllData As Range
    
    With [Master]
        
        Set rAllData = .Range("Header_LastName").CurrentRegion
        
Debug.Print "rAllData = " & rAllData.Address
Debug.Print "Range(Header_Event).Address = " & .Range("Header_Event").Address
Debug.Print "Range(Header_ShiftStart).Address = " & .Range("Header_ShiftStart").Address
Debug.Print "Range(Header_StartTime).Address = " & .Range("Header_StartTime").Address
Debug.Print "Range(Header_LastName).Address = " & .Range("Header_LastName").Address
    
        With .Sort
        
            .Header = xlYes
            
            .SortFields.Clear

            .SortFields.Add Key:=Range("Header_Event").Offset(1), Order:=xlAscending
            .SortFields.Add Key:=Range("Header_ShiftStart").Offset(1), Order:=xlAscending
            .SortFields.Add Key:=Range("Header_StartTime").Offset(1), Order:=xlAscending
            .SortFields.Add Key:=Range("Header_LastName").Offset(1), Order:=xlAscending
        
            .SetRange rAllData
            .Apply
        
        End With '.Sort
                
        Application.CalculateFull
        
    End With

End Sub

rAllData = $D$7:$V$107
Range(Header_Event).Address = $G$7
Range(Header_ShiftStart).Address = $J$7
Range(Header_StartTime).Address = $M$7
Range(Header_LastName).Address = $F$7
 

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So, there was an array in a hidden column. I Used offset to put that array into the data table inadvertently.
 
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