So, in other words my situation is hopeless?

Tom Ogilvy Wrote:
The code does nothing with pagebreaks. You would need to determin
where
the pagebreaks are on each insertion and then see how many rows need t
be
inserted if there is a problem. This would need to be done from th
first
row to the last row and you are moving from the last to the first, s
you
would probably need to do that in a separate loop.

You could just put in a pagebreak at each change in part number. D
you
want to do that?

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Regards,
Tom Ogilvy


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Not if someone wants to invest the time to write a solution for you. It
certainly isn't a trivial undertaking.

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Tom Ogilvy

"DKY" wrote in message
...

So, in other words my situation is hopeless?

Tom Ogilvy Wrote:
The code does nothing with pagebreaks. You would need to determine
where
the pagebreaks are on each insertion and then see how many rows need to
be
inserted if there is a problem. This would need to be done from the
first
row to the last row and you are moving from the last to the first, so
you
would probably need to do that in a separate loop.

You could just put in a pagebreak at each change in part number. Do
you
want to do that?

--
Regards,
Tom Ogilvy



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Now that I'm revisiting this with a little more experience and knowledge
I still don't know how I would do this. Here's what I'm thinking the
process might be in english.

For Pagebreak = Pagebreak +1
If the page break does not have a blank row under it Then
move up the rows until you find a blank row and insert spaces until the
data has moved below the page break and there is a blank row under the
page break.
End If
Next Pagebreak

Problem is, I have no idea how to code this further then what I've done
above. How do I get the pagebreak in code?


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Or, maybe.... Can I do something like this?

Clear all page breaks

Put page break at 40 rows down (or some number)
If row under the page break has a value then
place page break at 40 rows - 1
End i

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::bump::::bump::


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This must be a really tough one, I search and search and can't find
anything on it.


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