Solved: Use Offset to return a value

[YellowLabPro]

I cannot get this to work:
[vba]
If Not IsError(Application.Match(c.Value, Array("SNBD", "SNBB", "SNBT", "SNBJ", "SNBP", "SNBGG", "SNBH", "SNBGL", "SNBPK", "SHOES", "HAT", "BEAN", "TEE"), 0)) _
And Len(Range(c.Offset(0, 55)).Value) < 1 Then
[/vba]

Method of Range Failed
It has to do w/
[vba]Range(c.Offset(0, 55)).Value[/vba]
I can get this to work:
[vba]MsgBox Len(Range("BC6").Value)[/vba]

Posted Here originally:
http://www.mrexcel.com/board2/viewtopic.php?t=295928

[Oorang]

Is there a valid address string in the cell that c.Offset(0,55) evalutates to?

[YellowLabPro]

Is there a valid address string in the cell that c.Offset(0,55) evalutates to?

Hi Aaron,
Sorry- could you explain what you mean by valid address string?

[Oorang]

Well, this code segment:
[vba]Range(c.Offset(0, 55)).Value [/vba]Would evaluate like this:
Find range c evaluates to (we'll say C10). Go over 55 columns, (BF10), and get the value in the cell (because Value is the default property of the range object). If BF10 has something that is not an address, say "foo", then you end up with:
[vba]Range("foo").Value [/vba]

[YellowLabPro]

I have solved:
I was using Range before c.offset which it did not like. Why I don't know.

Aaron,
Thanks. That is an interesting, I had not considered that point. At the moment I was just looking if the lenght was greater than 0. My problem was as I remarked above was I hade the Range property, which I thought would be required.

[Paul_Hossler]

Hi

Instead of

Len(Range(c.Offset(0, 55)).Value) < 1

what happens if you use

Len(c.Offset(0, 55).Value) < 1 ??

Paul

[YellowLabPro]

Thanks Paul,
That is what I did use....