I am executing the following statement in VBA:

SetPtr = "Part 1"
Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False).Activate

And I get the following message:
Runtime Error 91:
Object Variable or With Block Variable not set.

What does this mean?

Thanks,
Mac Lingo
Berkeley, CA

If "Part 1" wasn't found, then you're trying to activate something that doesn't
exist--and it blows up.

The more usual approach:

dim FoundCell as range
SetPtr = "Part 1"
set foundcell = Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False)

if foundcell is nothing then
msgbox "Not found"
else
foundcell.activate
end if



Mac Lingo wrote:

I am executing the following statement in VBA:

SetPtr = "Part 1"
Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False).Activate

And I get the following message:
Runtime Error 91:
Object Variable or With Block Variable not set.

What does this mean?

Thanks,
Mac Lingo
Berkeley, CA

--

Dave Peterson

It means that the text "Part 1" was not found. You can avoid this error
by setting a range object when the value is found eg:

Dim SetPtr As String
Dim fndRng As Range
SetPtr = "Part 1"
Set fndRng = Cells.Find(What:=SetPtr)
If Not fndRng Is Nothing Then
fndRng.Select 'or whatever
End If

Hope this helps
Rowan

Mac Lingo wrote:
I am executing the following statement in VBA:

SetPtr = "Part 1"
Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False).Activate

And I get the following message:
Runtime Error 91:
Object Variable or With Block Variable not set.

What does this mean?

Thanks,
Mac Lingo
Berkeley, CA

Hi Mac Lingo,

The find method returns a range object which needs to be set.

Try something like:

Dim SetPtr As String
Dim rng As Range

SetPtr = "Part 1"

Set rng = Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False)
If Not rng Is Nothing Then
'Your code
End If

---
Regards,
Norman



"Mac Lingo" wrote in message
m...
I am executing the following statement in VBA:

SetPtr = "Part 1"
Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False).Activate

And I get the following message:
Runtime Error 91:
Object Variable or With Block Variable not set.

What does this mean?

Thanks,
Mac Lingo
Berkeley, CA

The find method returns a range object which needs to be set.

Should read:

The find method returns a range object which needs to be set to a range
variable which can be tested; or else you need to use an error handler to
prevent the error which results if the search string is not located.

The code was fine, the explanation was prematurely truncated!

---
Regards,
Norman



"Norman Jones" wrote in message
...
Hi Mac Lingo,

The find method returns a range object which needs to be set.

Try something like:

Dim SetPtr As String
Dim rng As Range

SetPtr = "Part 1"

Set rng = Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False)
If Not rng Is Nothing Then
'Your code
End If

---
Regards,
Norman



"Mac Lingo" wrote in message
m...
I am executing the following statement in VBA:

SetPtr = "Part 1"
Cells.Find(What:=SetPtr, _
After:=ActiveCell, _
LookIn:=xlValues, _
LookAt:=xlPart, _
SearchOrder:=xlByRows, _
SearchDirection:=xlNext, _
MatchCase:=False, _
SearchFormat:=False).Activate

And I get the following message:
Runtime Error 91:
Object Variable or With Block Variable not set.

What does this mean?

Thanks,
Mac Lingo
Berkeley, CA