object required error

[thechipper1982]

I can not figure out why I am getting this error. Any help would be great!

Here is my code:

Sub SaveOrder()

Sheets("ORDERS").Select
Range("A6").Select

Dim rownum As Long
Dim First As String
Dim Last As String
Dim Invoice As String
Dim MyDate As String
Dim findinvoice As String

Do While Not IsEmpty(ActiveCell)
ActiveCell.Offset(1, 0).Select
Loop
     
rownum = ActiveCell.Row
    
        Application.EnableEvents = False
        On Error Resume Next
      
  With Sheets("ORDERS")
      
            Set findinvoice = .Range("U:U").Find(Invoice, LookIn:=xlValues, LookAt:=xlWhole)

I get the error on the very last line with my string findinvoice.

I can not even get any part of the code to work, if I try to step into it, it stops at the very top and highlights findinvoice as the problem.

Not sure what to do.

chip.

 

[AlphaFrog]

Dim findinvoice As Range

The .Find method returns a range object.

 

[Kenneth Hobson]

Welcome to the forum!

AlphaFrog spotted the problem.

 

[thechipper1982]

thanks for the help.

that makes sense.

so the Set will return a cell reference?

I am trying to search through column U for the value stored in "Invoice". And if it finds that value then I want to set "findinvoice" to the value of that cell.

if I just change findinvoice to Range, will that allow me to set the cell's value to what I find in the cell?

thanks for help!

chip.

 

[AlphaFrog]

This finds the cell that matches the Invoice value.

    With Sheets("ORDERS")
        Set findinvoice = .Range("U:U").Find(Invoice, LookIn:=xlValues, LookAt:=xlWhole)
    End With

findinvoice is that matched cell. If the match was found in say "U10", then findinvoice is cell U10.

Once findinvoice is Set, you can treat it just like any Range object. You can change the value of findinvoice with somthing like...

findinvoice.Value = "XYZ123"

 

[thechipper1982]

can I say something like:

findvalue.select
if activecell.value = Invoice then
yadda yadda yadda
end if

should this work?

thanks

chip.