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I have a document that needs to be filled out. To help the user input all the correct information that goes on the form, I built a user form. On enter, the information gets transferred to the form. To accomplish this I use this statements like the ones below:

ActiveDocument.Bookmarks("NAME").range.InsertBefore txtName
ActiveDocument.Bookmarks("ADDRESS").range.InsertBefore txtAddress

This works great because I have separate bookmarks for Name and Address. But, what if I just have one bookmarks called "DATA":

ActiveDocument.Bookmarks("DATA").range.InsertBefore txtNetWeight _
& vbTab & txtCostPerLbs & vbCrLf

After this gets put on the document, I ask the users if he needs to input data again. If he clicks on Yes, then:

ActiveDocument.Bookmarks("DATA").range.InsertBefore txtNetWeight _
& vbTab & txtCostPerLbs & vbCrLf

BUT, here's the problem. I need the second set of data to be inserted below the first line that is already on the document. What happens instead is that it gets inserted above the line that is already on the document.

Any ideas? :roll:

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Last Post by I_Byte
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Well, I'm a VB.NET developer, and I don't really know much about VBA with MS Word, but, why not try creating a new bookmark at the end of your inserted portion, and then insert at that new bookmark?

You could then have an unlimited amount of inserts without a problem.

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Thanks for the reply. I actually received some help yesterday from a different source, Martin Green's Office Tips, and my problem is now solved. You were correct in your suggestion. Here's what Mr. Green suggested:

Dim DataRange As Range
Set DataRange = ActiveDocument.Bookmarks("Data").Range
If DataRange = "" Then
DataRange = "Some Text"
Else
DataRange = DataRange & " " & "More Text"
End If
ActiveDocument.Bookmarks.Add "Data", DataRange

Where I have written "Some Text" and "More Text" you can substitute with you textbox names. My example puts a space between the insertions. You could puta line break instead:

DataRange = DataRange & vbCrLf & "More Text"

I think that would work :)

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